For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
Inverse Functions
 A function does a transformation on an input. But what if there was some way to reverse that transformation? This is the idea of an inverse function: a way to reverse a transformation and get back to our original input.
 To help us understand this idea, imagine a factory where if you give them a pile of parts, they'll make you a car. Now imagine another factory just down the road, where if you give them that car, they'll give you back the original pile of parts you started with. There is one process, but there is also an inverse process that gets you back to where you started. If you follow one process with the other, nothing happens.
 It's important to note that not all functions have inverses. Some types of transformation cannot be undone. If the information about what we started with is permanently destroyed by the transformation, it cannot be reversed.
 A function has an inverse if the function is onetoone: for any a, b in the domain of f where a ≠ b, then f(a) ≠ f(b). Different inputs produce different outputs.
 We can see this property in the graph of a function with the Horizontal Line Test. If a function is onetoone, it is impossible to draw a horizontal line somewhere such that it will intersect the graph twice (or more).
 Given some function f that is onetoone, there exists an inverse function, f^{−1}, such that for all x in the domain of f,
In other words, when f^{−1} operates on the output of f, it gives the original input that went into f. [Caution: f^{−1} means the inverse of f, not [1/f]. In general, f^{−1} ≠ [1/f].]f^{−1} ⎛
⎝f(x) ⎞
⎠= x.  We can figure out the domain and range for f^{−1} by looking at f. Since the set of all outputs is the range of f, and f^{−1} can take any output of f, the domain of f^{−1} is the range of f. Likewise, f^{−1} can output all possible inputs for f, so the range of f^{−1} is the domain of f.
 The inverse of f^{−1} is simply f. This makes intuitive sense: if you do the opposite of an opposite, you end up doing the original thing.
 Visually, f^{−1} is the mirror of f over y=x. This is because f^{−1} swaps the outputs and inputs from f, which is the same thing as swapping x and y by mirroring over y=x.
 To find the inverse to a function, we effectively need a way to "reverse" the function. This can be a little bit confusing at first, so here is a stepbystep guide for finding inverse functions.
1. Check function is onetoone; f(x) = x^{3}+1
2. Swap f(x) for y; y = x^{3}+1
3. Interchange x and y; x = y^{3}+1
4. Solve for y; y = ^{3}√{x−1}
5. Replace y with f^{−1}(x); f^{−1}(x) = ^{3}√{x−1}  While this method will produce the inverse if followed correctly, it is not perfect. Notice that in steps #2 and #3 above, the equations are completely different, yet they still use the same x and y. Technically, it is not possible for x and y to fulfill both of these equations at the same time. What's really happening is that when we swap in #3, we're actually creating a new, different y. The first one stands in for f(x), but the second stands in for f^{−1}(x). This implicit difference between y's can be confusing, so be careful. Make a note on your paper where you swap x and y so you can see the switch to "inverse world".
 Taking inverses can be difficult: it's easy to make a mistake. This means it's important to check your work. By definition, f^{−1} ( f(x) ) = x. This means if you know what f^{−1}(x) and f(x) are, you can just compose them! If it really is the inverse, you'll get x. Furthermore, since we know f( f^{−1} (x) ) = x as well, you can compose them in either order when checking.
Inverse Functions

 A function is onetoone when different inputs always produce different outputs. In other words, there are no two inputs that produce the same output.
 For f(x), we see that there are cases when different inputs produce the same output. For example, f(apple) = f(apricot) = f(avocado) = a. Thus, because it is possible to have two distinctly different inputs result in the same output, the function is NOT onetoone.
 A function is onetoone when different inputs always produce different outputs. In other words, there are no two inputs that produce the same output.
 For g(x) to be onetoone, it must be that if we put in two different values for x, we will never get the same result.
 If we think about this carefully, we can realize that the only way to get the same output for g(x) is to start with the same x. For example, if x=3, we get g(3) = 9. But there is no other number we could possibly plug in other than x=3 to produce 9 as the output. [If we want to formally prove this, we can do it as follows: Let a and b be two numbers such that g(a) = g(b). Then 3a = 3b. Thus, by algebra, we have a=b. Therefore, if the output is the same, it must be that the inputs were the same, which proves that g(x) is onetoone. You don't need to formally prove this, though: you can just think about how the function works.]
 The Horizontal Line Test says that if a function is not onetoone, you can draw a horizontal line somewhere on it that will intersect the graph twice (or more).
 There are many places a horizontal line can be drawn that will cut the graph twice. Thus, the graph is not onetoone.
 The Horizontal Line Test says that if a function is not onetoone, you can draw a horizontal line somewhere on it that will intersect the graph twice (or more).
 There is nowhere on the graph that a horizontal line can be drawn that will cut the graph twice. Thus, f(x) is onetoone. [Near the center of the curve, it might look like f(x) is flat enough that you could cut it multiple times with a single horizontal line. This is not true, though. Notice that even when it looks fairly flat, it is still slightly sloped. It is not changing much, but it is still changing enough to pass the horizontal line test.]
 f^{−1} is the inverse function of f: it cancels out whatever f does. In general, f^{−1} ( f( x) ) = x. By having the inverse operate on a function, it gets us back to where we started. Thus f^{−1} ( f( 4) ) = 4.
 The inverse of f^{−1} is f. Thus, just like f^{−1} cancels out f, f will cancel out f^{−1}: f ( f^{−1} (x) )=x. By having a function operate on its inverse, it gets us back to where we started. Thus f ( f^{−1} (−27) ) = −27.
 The above canceling can occur multiple times:
f^{−1} ⎛
⎝f ⎛
⎝f^{−1} ⎛
⎝f(x) ⎞
⎠⎞
⎠⎞
⎠= f^{−1} ⎛
⎝f ⎛
⎝x ⎞
⎠⎞
⎠= x
 We already know the function is onetoone from the problem, so the next step is to replace f(x) with y:
y = 4x  Next, interchange the locations of x and y:
x = 4y  Solve for y from the new equation:
1 4x = y  Finally, replace y with f^{−1}(x):
f^{−1}(x) = 1 4x
 We already know the function is onetoone from the problem, so the next step is to replace f(x) with y:
y = x−3 x+3  Next, interchange the locations of x and y:
x = y−3 y+3  Solve for y from the new equation. This is kind of tricky: use the distributive property in reverse to pull out y once you have everything involving y on one side. xy + 3x = y−3 ⇒ xy−y = −3x −3 ⇒ y(x−1) = −3x−3
y = 3x+3 1−x  Finally, replace y with f^{−1}(x):
f^{−1}(x) = 3x+3 1−x
 Two functions f and f^{−1} are inverses when f^{−1}( f(x) ) = x or, equivalently, when f( f^{−1}(x) ) = x. This also means we can check to see if two functions are inverses by composing one with the other. Thus, if we can show f(g(x) ) = x or g ( f(x) ) = x, we have shown that they are inverses.
 This means we have two options. Let us show f(g(x) ) = x is true first:
Simplifying the above, we find that f(g(x) ) = x, and therefore f and g are inverses.f ⎛
⎝g(x) ⎞
⎠= f ⎛
⎝x−3 2⎞
⎠= 2 ⎛
⎝x−3 2⎞
⎠+3  Alternatively, we can show g ( f(x) ) = x is true:
Simplifying the above, we find that g ( f(x) ) = x, and therefore f and g are inverses.g ⎛
⎝f(x) ⎞
⎠= g ⎛
⎝2x+3 ⎞
⎠= (2x+3) −3 2
 Two functions f and f^{−1} are inverses when f^{−1}( f(x) ) = x or, equivalently, when f( f^{−1}(x) ) = x. This also means we can check to see if two functions are inverses by composing one with the other. Thus, if we can show f(g(x) ) = x or g ( f(x) ) = x, we have shown that they are inverses.
 This means we have two options. Let us show f(g(x) ) = x is true first:
Simplifying the above, we find that f(g(x) ) = x, and therefore f and g are inverses.f ⎛
⎝g(x) ⎞
⎠= f ⎛
⎜
⎝3⎛
√1 2x+6 ⎞
⎟
⎠= 2 ⎛
⎜
⎝3⎛
√1 2x+6 ⎞
⎟
⎠3
−12  Alternatively, we can show g ( f(x) ) = x is true:
Simplifying the above, we find that g ( f(x) ) = x, and therefore f and g are inverses.g ⎛
⎝f(x) ⎞
⎠= g ⎛
⎝2x^{3} −12 ⎞
⎠= 3⎛
√1 2(2x^{3} −12)+6
 The domain is the set of all values that the function can accept, while the range is the set of all values the function can possibly output.
 For f(x), it "breaks" when there is a negative in the square root, so the domain of f is x ≥ 5. The range of f is [0, ∞).
 To find f^{−1}, we follow the same steps we did in previous problems. Working it through, we get f^{−1} (x) = x^{2} + 5. [Check it by plugging one function into the other if you're not sure.]
 At first glance, we might think the domain of f^{−1}(x) is all numbers because x^{2} +5 never "breaks". However, we have to remember that f^{−1} is the inverse of f: it can only reverse values that could possibly come out of f. Thus, the domain of f^{−1} is the range of f: [0, ∞). Similarly, the range of f^{−1} is the domain of f: x ≥ 5 (Alternately, we can use the domain of f^{−1} to figure out what its range must be).
f^{−1}(x) = x^{2} +5
Domain of f^{−1}: [0, ∞) Range of f^{−1}: [5, ∞)
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Inverse Functions
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro 0:00
 Introduction 0:04
 Analogy by picture 1:10
 How to Denote the inverse
 What Comes out of the Inverse
 Requirement for Reversing 2:02
 The Basketball Factory
 The Importance of Information
 OnetoOne 4:04
 Requirement for Reversibility
 When a Function has an Inverse
 OnetoOne
 Not OnetoOne
 Not a Function
 Horizontal Line Test 7:01
 How to the test Works
 OnetoOne
 Not OnetoOne
 Definition: Inverse Function 9:12
 Formal Definition
 Caution to Students
 Domain and Range 11:12
 Finding the Range of the Function Inverse
 Finding the Domain of the Function Inverse
 Inverse of an Inverse 13:09
 Its just x!
 Proof
 Graphical Interpretation 17:07
 Horizontal Line Test
 Graph of the Inverse
 Swapping Inputs and Outputs to Draw Inverses
 How to Find the Inverse 21:03
 What We Are Looking For
 Reversing the Function
 A Method to Find Inverses 22:33
 Check Function is OnetoOne
 Swap f(x) for y
 Interchange x and y
 Solve for y
 Replace y with the inverse
 Some Comments 25:01
 Keeping Step 2 and 3 Straight
 Switching to Inverse
 Checking Inverses 28:52
 How to Check an Inverse
 Quick Example of How to Check
 Example 1 31:48
 Example 2 34:56
 Example 3 39:29
 Example 4 46:19
Precalculus with Limits Online Course
Transcription: Inverse Functions
Hiwelcome back to Educator.com.0000
Today, we are going to talk about inverse functions.0002
A function does a transformation on an input; we have talked about functions for a while now.0005
But what if there was some way to reverse that transformation?0009
This is the idea behind an inverse function: it is a way to reverse a transformation, reverse the process that another function is doing.0012
It is a way to get back to our original input.0021
By way of analogy, let's imagine a factory where, if you give them a pile of parts, they will make you a car.0024
Now, if you take that car down the road to this other factory, you can give them that car,0029
and they will give you back the original pile of parts you started with.0034
There is one factory where they make cars out of parts, but then there is a second factory0038
where they take cars and break them down into the original parts that were used to make them.0042
There is one process, but there is also an inverse process that gets you back to where you started.0046
If you follow one process with the other immediately, it ends up as if you haven't done anything.0051
If you bring the pile of parts to the first factory, and then take that car to the second factory,0055
and they give you back the pile of parts, it is like you just started with a pile of parts and didn't do anything to it.0059
This is the idea behind an inverse function: it reverses a processit reverses a transformation and gets you back to where you started.0063
We have used the analogy of a function as a machine before;0071
and it is a good image for being able to get across what is going on with inverse functions, as well.0073
So, a function machine takes inputs, and it transforms them to outputs by some rule.0079
So, what we are used to is: we plug in x into the function f, and it gives out f acting on x, f(x).0084
Now, we could plug it into another one; we could plug it into an inverse machine, an inverse to f;0092
and that would be called "f inverse," the f with the little 1 in the corner.0097
f^{1} denotes the inverse of f; we call it "f inverse."0103
We plug f(x) into f^{1}, into that machine; we get right back to our original x.0108
It is as if we hadn't done anything; the first machine does something, but then the second machine0114
reverses that process and gets us back to where we started.0119
So, is there a requirement for reversingcan we make an inverse out of all functions out there?0123
No; let's see why by analogy first.0129
Imagine a factory where, if you give them a pile of wood or a pile of metal, they give you a basketball in return.0133
The basketball is the exact same, whether you started with wood or metal; it is always the exact same basketball.0139
It doesn't matter what you gave them; it is just a basketball.0145
Now, let's say you walk down the road to another factory; you give them that basketball;0148
you tell them to reverse the process; and then you walk awayyou give them no other information.0152
Can that factory take a basketball and transform it back into a pile of wood or a pile of metal, if all they have is a basketball and no other information?0156
No, they have no idea what you started with.0164
Maybe they have wood; maybe they have metal; but the point is that they have no way0168
to be able to know which one they are supposed to give you at this point.0172
They don't have the information; the only person who has the information is you, when you brought the original wood pile,0175
or brought the original metal pilebecause all they have is the basketball, and the basketball could indicate wood, or it could indicate metal.0180
They have no way to know what you started with; there is no way to figure it out.0188
The information about what you originally had has been destroyed (although you would know it, because you brought it to the factory).0191
But assuming you forgot, then the information has been destroyedno one has the information anymore.0197
Another way to think about it would be if you took a piece of paper, and you burned that piece of paper.0202
You would be left with a pile of ashes.0206
Now, someone could come along and think, "Oh, a pile of ashesit used to be a piece of paper."0208
But if you take two pieces of paper, and you write two completely different things on the two pieces of paper,0212
and then you burn the two of them, a person could come along and think, "Two piles of ashes..."0217
And they would know it was paper, but they wouldn't know what was written on them.0221
They wouldn't be able to get that information back; the information has been destroyed.0224
They know it was paper, but they don't know what was written on the paper.0227
The basketball one...you have given them wood; you have given them metal; you get the same thing.0231
The information about what you started with has been lost; the information has been destroyed,0235
unless you come along and also say, "Oh, by the way, that basketball came from ____."0239
The issue, in this scenario, is that we have two inputs providing the same outputwhether it is metal or wood, you get a basketball.0245
So, if we try to have a reverse on that, we have no way to know which one to go back to.0252
We don't know if we are going to go back to metal; we don't know if we are going to go back to wood,0256
because we don't know what the basketball is representing.0259
So, to be a reversible processfor it to be possible to reverse somethingthe process has to have a different output for every input.0262
If you give them metal, they have to give you one color of basketball; and if you give them wood, they give you a totally different color of basketball.0271
Then, the second factory would say, "Oh, that is a wood basketball" or "Oh, that is a metal basketball."0277
And they would be able to know what to do at that point.0281
So, for a function f to have an inverse, it has to be that, for any a and b in the domain, any a or b that we could use in f normally,0284
where a is not the same thing as b (where a and b are distinct from each otherthey are different),0291
then f(a) is different than f(b); f(a) does not equal f(b).0297
So, if a and b are different, then the function's outputs on a and b are different, as well.0302
So, different inputs going into a function have to produce different outputs; we call this property onetoone.0308
If this function has a property where whatever you put in, as long as it is different from something else going in,0316
it means the two things will be different, that is called onetoone: different inputs produce different outputs.0323
You give them metal; you get one color of basketball; you give them woodyou get a different color of basketball.0328
Here are some examples, so we can see this in a diagram.0334
Here is an example that is onetoone: a goes to 2; b goes to 1; c goes to 3.0336
They each go to different things: different inputs each have different places that they end up going.0344
Something that is not onetoone: a goes to 1 and b goes to 1.0350
It doesn't matter that c goes to 3, because a and b have both gone to the same thing, so they have different inputs that are producing the exact same output.0354
a and b are different things, but they both produce a 1; so it is not onetoone.0365
We have that copy; we are putting in wood, and putting in metal, and we get basketball in both cases.0369
And then, finally, just to remind us: this one over here (hopefully you remember this) is not a function.0373
And it is not a function, because b is able to produce two outputs at once; and that is something that is not allowed for a function.0378
If a function takes in one input, it is only allowed to produce one output; it can't produce multiple outputs from a single input.0384
So, why do we call it onetoonewhy are we using this word, "onetoone"?0391
Well, we can think of it as being because a has one partner, and b has one partner, and c has one partner.0395
Everybody gets a partner, and nobody has shared partners; everybody gets their partner, and that partner is theirs.0402
They don't have to share it with anybody else.0408
It is onetoone: this thing is matched to this thing, and there is nobody else who is going to match up to that one: onetoone.0410
All right, how can we test for this?0417
One way to test for this, to test if a function is onetoone: we know, if we are going to be onetoone,0420
that every input must have a unique output; that was what it meant to be onetoone.0425
If we have different inputs, we have different outputs.0429
So, if we draw a horizontal line on the graph, it can intersect the graph only once, or not at all.0432
Remember, if we have some picture on a graphlike if we have this pointthen what that means0438
is that this is the input, and this over here is the output.0443
We make it a point: (input,output); that is why it is (x,f(x)).0449
If it is f(x) = x^{2}, then we plug in 3, and we get (3,9), (3,3^{2}).0458
The input is our horizontal location; the output is our vertical location.0465
The horizontal line test is a way to test if the function has the same output for multiple inputs.0469
We draw a horizontal line across, because wherever an output hits the graph, we know that there must be an input directly below it.0477
If a function is not onetoone, you will be able to draw a horizontal line that will intersect it twice, or maybe even more.0485
Let's look at some examples: first, here is one that is onetoone, because whenever we draw0490
any sort of horizontal line, it is only going to cut it once.0496
The only place that might seem a little confusing is if we draw it near here.0499
It might make you think, "Well, doesn't it look like those are stacked?"0503
Well, remember, we can't draw perfectly what is being represented by the mathematics.0506
We have to give our lines some thickness; in reality, the line is infinitely thin.0511
So, while it looks like they are kind of getting stacked, it is actually still moving through that zone; it is not constant.0517
It is increasing just a little bit, but it isn't constant.0522
Let's look at one that is not onetoone: over here, this horizontal line (or many horizontal lines that we could make)0525
it cuts it in two places; so we know that, here and here, there are two inputs.0532
We can produce the same output from two different locations.0537
We have two inputs making one output; so that means we are not onetoone,0541
because this one is partner to that height, but this one is also partner to this height.0545
So, we are not onetoone, because we have to share an output.0550
Now that we have all of these ideas in mind, we are finally ready to define an inverse function; we can really talk about them and sink our teeth into them.0554
Given some function f that is onetoone (it has to be onetoone for this to happen),0561
there exists and inverse function, which we denote as f^{1} such that, for all x in the domain of f,0566
any x that could normally go into f, any value that could normally be input,0574
f^{1} acting on f(x) becomes just x.0578
So, we have f acting on x like normal; and then, f^{1} acts on that whole thing.0583
And it breaks the action that was done by f and returns us back to our original input.0591
In other words, when f^{1} operates on the output of f, it gives the original input that went into f.0596
Caution: I want to warn you about something: f^{1} means the inverse of f, not 1/f.0604
This can be confusing, because, if you have taken algebra and remember your exponents0612
(you might have forgotten them, but we will talk about them later in this course), 1 can mean a reciprocal for numbers.0617
So, 7^{1} becomes 1/7; x^{1} becomes 1/x.0622
But f is not "to the 1"; it is just a symbol that says inverse"This is a function inverse."0630
So, in general, for the most part, f^{1}, "f inverse," is not the same thing as 1/f.0636
The inverse of f is normally not the reciprocal of f, 1/f.0643
This exponent thing, where 7^{1} is 1/7, is not the case for functions.0649
On a function, the 1 does not represent an exponent; it is not an exponent.0656
But it instead tells us function inverse; it is a way of saying, "This is an inverse function"; that is what it is telling us, not "flip it to the reciprocal."0661
How can we get domain and range for f^{1}?0673
We can figure these out by looking at f; remember, the set of all outputs from f is its range.0675
The things that x can get mapped to by f, what f is able to map x to, is the range of f.0687
The domain of f is everything that x can be, everything that we can plug into f.0695
And then, the range of f is everything that can come out of f.0701
Now, f^{1} has to be able to take any output of f.0704
It is not very good at reversing if there are some numbers that it is not allowed to reverse.0709
So, it has to be able to reverse anything from f.0713
If it is able to reverse anything from f, then that means the range of f has to be everything that we can put into f^{1}.0716
So, the domain of f^{1}, the things we can put into f^{1}, is the range of f.0723
The domain of f^{1} is the range of f.0728
Likewise, because f^{1} then breaks that f(x) and turns it back into original inputs,0732
we must be able to turn it back into all of the original inputs, because all of the original outputs have to be over here.0738
So, anything that we can make it to, we have to be able to make it back from.0745
So, since we are able to get back everywhere, that means that we can output all the possible inputs for f.0748
Since we can output all of the possible inputs, because we can reverse any of the processes,0755
then it must mean that we are able to get the range of f^{1} from the domain of f.0760
So, the domain of f tells us the range of f^{1}, what we are allowed to output with f^{1}.0767
And the domain of f^{1} tells us the range of f.0772
So, the domain of f^{1} is the things that f is able to output; and the range of f^{1},0776
the things that f^{1} is able to output, is what we can put into f, the domain of f.0784
This idea is going to let us prove something later on.0790
Now we can get to that proofthe inverse of an inverse: what is the inverse of an inverse?0793
In symbols, what is (f^{1})^{1}what do we do if we are going to take the inverse of something that is already doing inverses?0797
Now, it might seem a little surprising, but it turns out that the inverse of f^{1} is just f.0806
The inverse of f^{1} is f; it seems a little surprising, maybe, but it makes a sort of intuitive sense.0812
If you do the opposite of an opposite, you get to the original thing.0820
If you do an action, but then you are going to do the opposite of that action,0824
but then you do the opposite of the opposite of the action, then you must be back at your original action.0828
So, we might be able to believe this on an intuitive level; it makes sense, intuitively, that two opposites gets us back to where we started.0833
But let's see a proof of this fact, formally; let's see it in formal mathematics.0840
So, how do we get this started? Well, by definition, f^{1} is the function where,0845
for any x, f^{1} acting on f(x) is going to just give us our original x.0849
If f acts on an input, and then f^{1} comes and acts on that, we get back to our original input.0856
Now, consider (f^{1})^{1}: by this definition of the way inverses work, it must be that f inverse, inverse,0862
when it acts on the thing that it is an inverse of...f inverse, inverse, is an inverse of f inverse...0869
I know it is complicated to say...but this one right here is going to be the opposite action of f^{1}.0874
So, if we take any y (don't get too worried about x and y; remember, they are just placeholders for inputs),0881
similarly, for any y, (f^{1})^{1}, acting on f^{1}(y),0887
is going to just get us right back to our original y.0892
It is the same structure as what is going on here with f^{1}(f(x)) = x: we are just reversing a process.0895
So, it doesn't matter that one of the processes is already a reversed process, because we are reversing this other reversed process.0901
So, we get back to our original thing.0907
Now, we know that y is in the domain of f^{1}, because we are allowed to put it into f^{1}.0910
It is allowed to go into f^{1}; now, we know, from our thing that we were just talking about,0917
that the domain of f^{1} is the range of f; so there has to be...0922
If f^{1} is the range of f (the domain of f^{1} is the range of f),0930
if you are in the range, then that means that there is something out there that can produce this.0940
That means that, if you are in the range of f, there must be some x in the domain of f;0944
there has to be some way to get to that place in the range, so that f(x) is equal to y.0949
There is some x out there in the domain of our original f, that f(x) is equal to y.0954
So now, we have what we need: we can use this f(x) = y, and we can just plug it in right here and here.0960
We will plug it in for the two y's up here, and we will see what happens.0967
Thus, f inverse, inverse, acting on f^{1}(f(x))...0970
because remember, we know that there has to be some way to get an x such that f(x) = y,0974
because of this business about domain and range; so we plug that in here, and we plug that in here.0979
And we have that f inverse, inverse, on f inverse of f of x, must be equal to this over here on the right, as well.0985
We are just doing substitution.0993
But we know, by the definition of f^{1}(f(x)), that this just turns out to be x.0995
This whole thing right here just comes down to xit simplifies right out to x.1002
So, it must be the case that f inverse, inverse, of x is the same thing as f(x).1007
If f(x) is the exact same thing as f inverse, inverse of x, it must be the case that (f^{1})^{1} is just the same thing as f.1012
And our proof is finished; great.1023
How can we interpret this graphically?there is a great way to interpret inverses through graphs.1028
First, let's consider f(x) = x^{3} + 1.1032
Now, we know that this one has to be onetoone, because it passes the horizontal line test.1035
We come along and try to cut this with any horizontal line; they are only going to be able to cut in one place.1040
Even here, where we have sort of seemed to flatten out, it is still moving, because we know it is x^{3} + 1.1045
And it never actually stops going up; it just slows down how fast it is going up.1051
And our lines have to have thickness, so while it kind of looks like they are stacked, they are not really.1055
So, we see that it passed the horizontal line test; so it must be onetoone.1059
If it is onetoone, we know it has to have an inverse; that is how we talked about this, right from the beginning.1064
Now, notice that the graph, any graph, is made up of the points (x,f(x)).1069
We talked about this before: 0 gets mapped to 1 when we plug it in as f(0) = 1; so that gives us the point (0,1).1073
That is how we make up our original graph for f.1082
Now, the graph of f^{1} will swap these coordinates.1084
It takes in outputs and gives out inputs, in a way; so its input will swap these two things.1088
It takes in f(x), and it gives out x; so the points of f^{1} will be the reverse of what we had for the points of the other one.1096
So, (f(x),x) is what we get for f^{1}.1105
Now, visually, what that means is that f^{1} is going to be the mirror of y = x; and that is our line right here, y = x.1109
Why is that the case? Well, look: we swap x and y coordinates if we go across this,1117
because (3,0) swaps to (0,3); if we are going over y = x, if we are mirroring across this, we will swap the locations.1122
And so, if we mirror over y = x, we are going to swap x and y, y and x; we will swap the order of our points,1138
because y = x is sort of a way of saying, "Let's pretend for today that I am you and you are me."1145
y is going to pretend that it is x, and x is going to pretend that it is y.1153
They are sort of swapping places when we do a mirror over it.1156
So, that means our picture, mirroring f over y = x...we get the graph of f^{1}.1159
So, we look at this; we mirror over it; and we get places like this.1164
All right, we see how we are just sort of bouncing across it.1171
And this is going to happen with any of our inverses graphically.1179
So, any time f^{1} is being looked at, we know it is going to be a bounce, a reflection through, a mirror over;1183
it is going to be symmetric to f with respect to the line y = x.1190
Since f^{1} is swapping outputs and inputs, it is going to be sort of reversing the placements of these.1193
So, the graph of f^{1} will always be symmetric to f, with respect to the line y = x.1201
It will bounce across, because when you bounce across y = x, you swap your coordinates.1206
Now, there are many ways to say this; I am saying "bounce across"; that is not really formal.1211
But we can formally say that it is symmetric to f, with respect to the line y = x.1215
You could also say that it reflects through y = x, or it reflects over y = x.1220
You could also say that it mirrors over, or it mirrors through, y = x.1224
There are many ways to say it; but in all of these things, the same idea is that we are going to bounce across,1227
and that that point will now show up at that same distance here.1232
So, let's see what it looks like: we bring them in, and indeed, they pop into those places.1235
They pop into being a nice, symmetrictotheline, y = x; and that makes sense.1240
We replaced the inputs with outputs and outputs with inputs; they have swapped locations.1247
We look at this one here, and the point (3,0) on the inverse is connected to (0,3) on the original functionthe same sort of thing on both of them.1252
All right, so we have talked a lot about what is going on; we have a really great understanding of the mechanics behind an inverse.1264
But how do we actually find an inverse?1270
Now that we understand them, we are ready to actually go and find them.1273
How do we turn an algebraic function like f(x) = x^{3} + 1 into a formula?1276
Before we do this formula for f^{1}, consider that f^{1} is taking the output of f(x); and it is transforming that into x.1282
To find a formula for f^{1}, we want a formula that gives x, if we know f(x).1290
Normally, f(x) = x^{3} + 1, for examplenormally we have x.1298
We know x, and from that, we get our f(x); you plug in an x into a function, and it gives out f(x).1303
So, f^{1} is the reverse of that; we know f(x), and we want to get x out of it.1312
So, to be able to do this, we are solving f(x) = x^{3} + 1 in reverse.1318
f(x) = x^{3} + 1; well, we move that over: f(x)  1 = x^{3}; so now we have ^{3}√(f(x)  1) = x.1323
If we know what f(x) is, we can figure out that that is what the original x that did it is.1338
We are solving it in reverse; we have reversed the function.1342
As opposed to solving f(x) in terms of x, we are solving x in terms of f(x).1345
Now, this is a little bit of a confusing idea; so instead, I am going to show you a method to do this.1351
The idea of reversing is really what is behind inverses; it can be a little hard to understand what to do on a stepbystep basis.1357
We are normally used to solving for f(x) in terms of x, having f(x) just on its own on one side, and having a bunch of stuff involving x on the other side.1365
So, at this point, it might be a little bit confusing for you to try to do it the other way.1372
And it would work; but let's learn a method that makes some of that confusion go away, and do things we are more used to doing.1376
Here is one stepbystep guide for finding inverse functions.1382
The very first thing we have to do: we have to check that the function is onetoone.1384
It has to be onetoone for us to be able to find an inverse at all.1388
Now, f(x) = x^{3} + 1...we just saw its graph; remember, it looked something like this.1391
So, we already know that it passes the horizontal line test; it does a great job; it is a great function.1397
It is onetoone; greatwe have already passed that part for this.1401
Next, we swap f(x) for y; this is going to be a little bit easier for us in solving.1406
We are used to solving for y's; we are not really used to solving for f(x)'s; so this will make it a little bit less confusing.1410
We switch out f(x) for y; great; in the next step, we interchange the x and the y.1415
In this one, we have x in its normal place and y in its normal place.1423
What we do on step 3 is swap their places: y takes the place of all of the x's, and x takes the place of y.1427
We swap x and y, interchange x and y; every time you had an x in step 2, you are now going to have a y;1436
every time you had a y (which is probably just the one time, since it was from a function), you are going to now have an x.1443
That is how we are doing this step that is the reversing step.1448
Solve for y: at this point, we have x = y^{3} + 1; so if x = y^{3} + 1, we solve for it.1453
We just move that 1 over: x  1 = y^{3}; and we have ^{3}√(x  1) = y.1460
And you will notice that this actually looks pretty much the exact same as what we just did on the previous slide1469
but perhaps a little less confusing, because it is what we are used to seeing.1473
So, we have y = ^{3}√(x  1); and finally, just like we replaced f(x) with y, we now do a reverse replacement.1476
But we are now going to f^{1}; so y now becomes f^{1}(x).1486
f^{1}(x) is equal to ^{3}√(x  1); f^{1}(input) = ^{3}√(input  1).1492
Great; now, while this method will produce the inverse if followed correctly, it is not perfect.1500
Now, remember steps #2 and #3; in that, we had to swap f(x) for y, and then we were told to interchange x and y.1507
Remember, they swapped places; now notice, these equations are completely different.1515
They are totally, totally different from one another; yet they are still using the same x and y.1520
Technically, it is not possible to have both of these equations be true with the same x and y.1530
x and y can't possibly fulfill both of these equations at the same time, because they are completely different equations.1534
So, what is going on here? When we swap in step #3, we are really creating a new, different y.1541
When we have "swap f(x) for y," it is really red y or something here.1548
But then, when we do the interchange, it becomes a totally different color of y; it becomes like blue y here.1553
So, we are creating a new, different y; the y when we first swap is different than the meaning of the second y.1560
The swapping y is a different y from our first time that we replaced f(x).1567
The first one is standing in for f(x); that was our red y.1572
And then, the second one stands in for f^{1}(x); that is really taking the place there.1578
This implicit difference between y's can be confusing; so be careful.1584
I would recommend making a note on your paper; make a note when you are working that says where you swap.1588
Use a note to see that swap of x and y, so that you can see the switch over to this inverse world,1593
where you are now in an inverse world, and you can solve for an answer.1599
This is a bit confusing; so why are we learning this method, if it has this hidden, confusing1603
implicit difference, when we really think about what is going on?1609
In short, the reason we are doing this is because everybody else does.1612
That is not because it is perfect; it is because everyone else out there pretty much learns this method for solving inverses.1616
Most textbooks, and almost all of the teachers out there, teach this method.1623
So, it is important to learn, not because it is absolutely, perfectly correct, but because it is standard1627
so that you can talk to other people, and talk about inverses, and they will understand1632
what you are talking about, because they are doing the same method that you are doing.1635
If you do something different, they might get confused.1639
If they are really clever, or they really understand what is going on, they will think, "Oh, yes, that makes perfect sense."1641
But we want to go with the standard method, so that other people will understand what we are doing.1645
And if we are taking a course at the same time as we are watching this course,1649
the teacher will think that is correct, as opposed to being confused by what you are doing and marking your grade down.1653
But the important idea here, the really important idea inside of this thing, that is confusing, is the reversal.1657
That is what the moment is all aboutthat moment between #2 and #3the #3 step where we reverse, and we create this new y.1663
We reverse the places; instead of solving for an output, we are solving for input.1671
We are reversing the places, so we can do this directly; I did that with f(x), where I did f(x) = y;1678
and I solved directly for if we know what f(x) is.1685
I'm sorry, f(x) equals stuff involving x; I solved it directly for f(x)...1687
We had f(x) = x^{3} + 1, and we figured out that it also is the same thing that the cube root of f(x)  1 is equal to x.1694
We figured that out; so there is this direct way of being able to do this.1706
We can do this directly; but lots of students find this difficult or confusing, so we have this method of swapping x and y.1709
And also, it has just become the standard way to do things; so it is good to practice this way, even though it is not absolutely perfect.1716
It is not a perfect method, but it does the job.1723
As long as you are careful and you pay attention to what you are doingyou closely follow its steps1725
you will be able to get the answer, and you will be able to find the inverse function.1729
Taking inverses can be difficult; it seemed a little bit confusing from what I have been saying so far.1733
And it is an easy one to make a mistake on; this means it is really important to check your work.1737
You really want to make sure that you check your work on this.1743
How do you do this? Well, remember: by definition, f^{1}(f(x)) is equal to x.1746
That means, if we know what f^{1} is (we have figured out its formula), and we know what f(x) is1752
(we were probably told f(x), we can just compose them.1756
We know how to compose them from our lesson Composite Functions.1760
If you didn't check out Composite Functions, you will have to watch that before you are able to compose them and do this check.1763
But if it is really the inverse, you will get x; if you compose f^{1} with f(x), it has to come out to be x,1768
because that is the definition of how we are creating this stuff, right from the beginning.1775
Furthermore, we also know that f^{1}, inverse, was just f;1779
so it also must be the case that f acting on f^{1}(x) will give us x, as well.1785
You can compose them in either order when you are doing a check; and you will end up being able to get it correct.1790
Let's see a quick example: for example, if f(x) = x^{3} + 1, and f^{1}(x) = ^{3}√(x  1)1796
(the ones we have been working with), how do we check this?1803
Well, let's start with f^{1}(f(x)); we compose this: we plug in f(x) = x^{3} + 1.1805
So, f^{1} acting on x^{3} + 1...now, remember, we are going to plug that into f^{1}(x).1814
But it is f^{1}(input); whatever is in the box just goes to the box over here.1821
So, it is going to be that f^{1} will become cube root...where does the box go?1825
x^{3} + 1...that is our box...minus 1; so the cube root of x^{3} + 1  1...1832
+ 1  1 cancels; the cube root of x^{3} equals x; greatthat checks out.1841
What about if we did it the other wayif we did it as f(f^{1}(x))?1847
Hopefully, this will work out, as well (and it will).1852
So, what is f^{1}(x)? f^{1}(x) is the cube root of x  1, so f(^{3}√(x  1))...1854
what is going to happen over here?we know that you plug in the box; you plug in the box.1864
So, f(^{3}√(x  1))...we are going to take that, and we are going to plug it in right here.1869
It is going to be ^{3}√(x  1), the quantity cubed, because it has to go in as the box; plus 1finish out that function.1875
The cube root, cubed...those are going to cancel each other; we will get x  1 + 1, which is just equal to x; and it checks out.1885
So, we can check it as f^{1}(f(x)) or f(f^{1}(x)); sometimes it might be easier for us to do it one way or the other.1895
We could also do both ways, if you want to check and be absolutely, doubly sure that we really got our work correct.1902
All right, let's move on to some examples.1907
Using these graphs for assistance, which of the following functions are onetoone?1909
The first one is f(x) = 1/x; we do the horizontal line testit is going to pass any high horizontal lines.1914
What about as we get lower? Well, we know that 1/x continues to moveit never freezes and becomes constant.1921
Does it ever cross this xaxis, though? No, it doesn't.1928
We haven't formally talked about asymptotes yet; we will talk about asymptotes in a later lesson.1932
But 1/x...as we go positive (f of a positive), 1 over a positive is going to also have to be positive.1936
So, it never crosses the xaxis; the same thing goes with the negativesf of a negative is going to be a negative.1945
So, when it goes to the left, it never manages to cross this xaxis; as it goes to the right, it never manages to cross this xaxis.1951
And it keeps changing; so the two things never cross over each other.1958
So, yes, this is onetoone.1961
What about the blue one, g(x) = x^{3}  2x^{2}  x + 1?1968
It is easy to say it fails: we cross lots of places in the middle here, and it is able to have multiple points at the same time.1973
So, any one of these hits here and here and here; there are three points that all give the same output of 0; so it fails the horizontal line test.1982
It is not onetoone.1992
Finally, (2x  1) and (x^{2} + 1); 2x  1 is just a line that is going to keep going on this way forever and ever and ever.1998
2x  1, when x is less than or equal to 1...this is from piecewise functions; if you haven't checked out piecewise functions, this might be a little confusing.2007
But hopefully, you have watched that lesson already.2013
2x  1 is x ≤ 1; it is just going to keep going on down and down and down, to the left and left and left.2015
And x^{2} + 1 is the right side of the parabola; if we plug in higher and higher numbers, it just keeps curving up and up and up to the right.2020
So, that means that we are never going to cross; the parabola is never going to double back and manage to touch itself again.2027
The parabola might eventually do this, but that part isn't on it.2033
And the line is never going to be able to go down to have itself crossed horizontally.2037
So, if we do any horizontal line crossing on this, it is never going to hit twice; so it is onetoone.2042
One thing I would like to make a special comment on: notice that right here there is an empty space.2055
There is this gap where it jumps; is that a problem for a horizontal line test?2060
No, it is not a problem at all, because the horizontal line test is allowed to hit no points, as well.2064
It is allowed to hit one point or zero points; in this case, if it goes through that gap, it hits no points; but that is OK.2070
We are only worried about having multiple inputs for the same output.2076
It is OK if there are no inputs to make an output; the important thing2080
is that there are no double sets of inputs that all make the same output.2082
Like, in the blue one, where we had multiple different places where we could plug in some number2087
plug in different numbers, but they would all produce zeroes.2092
All right, let's actually find an f^{1}: f(x) = 3x/(x + 3).2096
They told us, right from the beginning, that it is onetoone; so we can jump right to figuring it out: what is f^{1}(x)?2102
And then, after it, we need to check our answer.2108
OK, so what is f^{1}(x)? Remember all of our steps, one by one.2111
f(x) = 3x/(x + 3): they told us, right from the beginning, that it is onetoone, so we are already checked out.2115
We have already checked out the first one.2122
The next step: we swap y for f(x): y = 3x/(x + 3).2124
Now, that is not the important part of when we reverse, though; we reverse into inverse world.2131
So, here is when we go into inverse world; we reverse the place of x and y.2136
So now, it is x where y was, and it is 3y/(y + 3).2144
Now, we just go about this, and we solve it for y like we normally would.2151
Multiply both sides by y; we get x times (y + 3) equals 3y; let's distribute this out: xy + 3x...let's also move the 3y over, so + 3y = 0.2154
OK, at this point, we will pull out the y's from these two things; we will move them together, so we can see it a little bit easier at first.2170
xy + 3y + 3x = 0; let's subtract that 3x to move it over; 3x, 3x here.2176
So, then we will pull out the y's to the right; so we have x + 3, times quantity y, equals 3x.2186
Finally, we divide by that x + 3, and we get y = 3x/(x + 3).2195
And now, finally, we can plug in f^{1} for this y; so we plug it in, and we get f^{1}(x) = 3x/(x + 3).2202
Great; now, let's check and make sure that we got this right.2216
We check this in red; here is our checklet's check it by plugging f into f^{1}.2219
So, we want this to come out to be x; it should be x, if we got everything right.2232
So, f^{1}(f(x)); what is f(x)? f(x) is this; and here is something funny to notice.2239
Notice 3x/(x + 3); amazingly, it just so happens that for 3x/(x + 3), f(x) and f^{1}(x) are the exact same thingkind of impressive.2245
We plug this in; we have f^{1}(f(x)); f(x) is 3x/(x + 3); now, over here, we plug it in; what is in the box?2260
The box shows up here; the box shows up here; it shows up twice, so it is f^{1} on 3x/(x + 3).2274
It is going to be 3...what is in the box?...3x/(x + 3), over (3x/(x + 3)) + 3.2281
Great; so the first thing that is going to be confusing is that we have this x + 3, and we have this x + 3 here.2301
So, let's take that out by multiplying the whole thing by (x + 3)/(x + 3).2307
We can get away with that, because it is just the same thing as 1: (x + 3)/(x + 3) is just 1.2312
So, (x + 3)/(x + 3)...multiply that here; the (x + 3) will cancel out here and cancel out here.2317
But remember, it also has to distribute to the other part, because they are not connected through multiplication on that; they are connected through addition.2324
So, we have 3, 3x, over 3x plus 3 times x plus 3.2330
These two negatives cancel out; so we have 3 times 3x on the top, 3x plus 3(x + 3)...so we have 9x on the top,2341
divided by 3x plus 3x plus 9; 3x plus 3x...they cancel each other out; we have 9x/9.2350
9 over 9...those cancel out, and we have just x.2362
So, that checks outgreat, we have the answer.2367
All right, the next one: we have, this time, a piecewise function.2370
This is a little confusing: we didn't talk about this formula, but we will see how to do it.2375
f(x) = x + 1 when x < 0, and √x when x ≥ 0.2378
This is confusing; we don't know what to do about the different pieces of the piecewise function.2387
We don't know what to do about these two different categories: we have x < 0 and x ≥ 0.2393
We didn't learn that when we learned how to do inverses; but we could still figure out these two.2397
We could figure out what is the inverse of x + 1 and what is the inverse of √x.2402
We were told, explicitly, that this is onetoone; so we can go ahead and do this, and then we will think about it.2407
First, we will do inverses on these two rules; and then we will figure out how they fit togetherwhat are the categories for these two rules?2413
So, first, x + 1; we will have y = x + 1; we swap them, so we now get into our inverse world.2422
Swap their locations; we interchange them, and we are now at negative...sorry, not x; the negative does not swap.2439
We are at x = y + 1; we move the y over and move the x over; we get x + 1; so y = x + 1,2446
which is going to give us f^{1} for at least the first rule here.2458
Now, what about the other one?let's do that, as well.2464
So, y = √x; we go into inverse mode; we reverse their locations; and we are now at x = √y.2467
So, how do we solve for y? Well, we move this negative over: x = √y.2482
Square both sides; we get (x)^{2} = y; and then (x)^{2}...the negatives will cancel out, so we get just x^{2} = y.2487
And so, this is the inverse rule for this part.2498
Now, here is the part where we start thinking: we know that f^{1} is going to break into a piecewise function using these two different things.2502
y = x + 1...so it will be x + 1 for the first rule, and then x^{2} for the second rule.2513
But the question is that we don't know what the categories are.2520
How do we figure out what the categories are?2524
Well, remember: if f goes from its domain to its range, let's call that a to b, then f^{1} does the reverse of that.2526
f^{1} goes from b to a; it does the reverse.2543
What that means is that the domain...the thing that determined which rule we used...we need to do the range to determine which way to get back.2548
The range on these two rules...now we are back to using f, so range on f...for x + 1:2557
well, x + 1 was x < 0; that was the category, so it has to be within those.2571
So, what can it go to? Well, if we plug in a really big negative number, like, say, 100, we will get (100) + 1; so we get 101.2577
So, as long as we keep plugging in more and more negative numbers, we get bigger and bigger numbers.2584
We are able to get all the way out to positive infinity, as we are really far in negative numbers.2587
What is the lowest that we can get to? Well, we could get really close to 1, as we plug in 0.00000001.2592
We are really close to being to 1, so we can get right up to 1; but we can't actually touch it.2601
We have to exclude it, so we use parentheses.2605
So, the range for the first rule is this: x + 1 becomes this.2607
So, I will put a red dot on that, because that matches to this rule here.2614
Now, what about the range for the other rule?2619
The range on this rule is √x; it has x ≥ 0 as its domain; what are the numbers we can get out of this?2622
What is the largest number we can get out of it?2630
The largest number we can get out of it is actually 0; why?because, when we plug in any reasonably large positive number,2632
like, say, 100, then √100 is 10; so as we get bigger and bigger positive numbers that we plug in,2639
we actually get more and more negative.2649
So, we can actually go to any negative number we want; we can go all the way down to negative infinity.2651
Can we actually reach 0? Yes, we actually can reach 0, because it is greater than or equal to.2655
So, if we plug in x = 0, we get √0, which is just 0; and we put a bracket to indicate that we are actually allowed to do it.2660
This one is the range for √x, that rule; it is going to get a green dot on it, because it matches to the green rule.2668
That means that x + 1 is allowed to take in...what values? It is allowed to take in the range values.2677
It is allowed to do a reverse on anything that shows up in the range (1,∞).2684
Also notice: these two ranges, (1,∞) and (∞,0], don't have any intersection.2689
They don't overlap at all, so we don't have any worries about pulling from one versus pulling from the other.2695
They will never get in each other's way.2700
So, for this one, x + 1, if it is going to be allowed to go from 1 to infinity, then that means we can plug in anything into f^{1},2703
where x is greater than 1, which is to say input; it is not the same x that was up here.2711
It is now just saying "placeholderanything that we are plugging in."2719
What about x^{2}? Well, that was the green dotthat was allowed to go from negative infinity up to 0.2722
So, it is allowed to have x ≤ 0; it is allowed to go all the way up to negative infinity, but it can only just get to touching 0.2727
It is allowed to actually have 0, though; x > 1 is not actually allowed to touch 1, but it is able to get as close as it wants.2735
And there is our piecewise inverse function.2741
It is a little bit difficult, but if you think about it, you do each of the inverses, and then you think about2747
"How do I get the domain for the inverse? I get it from the domain of f becoming the range of our inverse,2750
and the range of our f becoming the domain of our inverse."2760
So, what the original function was able to output to is what the inverse is allowed to take in.2766
And that is how we figured out these rules, these categorieswhat the categories were for these two different transformations.2773
All right, the final example: f and g are onetoone functions; now, prove that f composed with g, inverse, is equal to g inverse composed with f inverse.2779
This might be a little daunting at first; these are weird symbols; we are not used to using these sorts of things.2791
So, if that is the case, let's remind ourselves: from composition, f composed with g, acting on x, is equal to f(g(x)).2795
Now, I said before: it makes things always, always, always easier to see it in that format.2806
What we want to show is that g^{1} composed with f^{1} (which would be g^{1}(f^{1}(x)))...2811
we want to show that this one here is an inverse to that one over there.2823
That is what we are trying to prove, that f composed with g inverse...2829
We know, by the definition of how this symbol works, by how inverses work...2833
f composed with g^{1}, acting on f composed with g, on x, is going to just leave us as if we had done nothing,2837
because we are putting an inverse on something.2845
So, we want to show that this means the exact same thing as this right here.2848
So, let's just try it out: we will set it up like this: f composed with g^{1}, acting on f composed with g, acting on x.2854
OK, so what does that become? Well, we know that f composed with g, acting on x, is the same thing as f(g(x)).2871
All right, what is f composed with g^{1}? Well, we know (from what we did over here)2881
that we can bring that into g^{1} acting on f^{1}, acting on whatever is going into it.2887
What is going into it here is this whole thing; so, it is going to be g^{1}, acting on f^{1}, acting on f, acting on g, acting on x.2891
And then, we close up all of those parentheses.2908
That is a little bit confusing; but we are seeing inverses right next to functions: f^{1} acting on f, acting on whatever is in there.2912
It just cancels out and gets us right back to what we originally had in there.2922
So, f^{1} acting on f...that cancels out, and we get g^{1}, acting on whatever was in there, which was g(x).2926
So, g^{1} acting on g(x)...the exact same thing: we get down to x; so we have proved it.2933
g^{1} composed with f^{1} is how we create f composed with g, inverse.2940
Great; we have proved it.2948
All right, I hope you have a much better idea of how inverses work at this point.2949
They can be a little bit confusing, but you have that method to be able to guide you through it.2952
Just follow it really carefully, stepbystep.2956
The danger is if you break from those steps and do something else; that is where you can make mistakes.2958
If you really understand what is going on, you don't even have to use that method.2963
But it really is the standard method, so it is a good idea to stick with it, just because it is what a lot of other people are used to using.2966
And you can find it in a lot of textbooks.2972
All right, we will see you at Educator.com latergoodbye!2974
1 answer
Last reply by: Professor SelhorstJones
Sat Feb 11, 2017 8:14 PM
Post by John Lins on February 11 at 08:07:15 PM
Please, help me with this question:
Find Laplace Transform of the following signal:
x2 (t)=(1(1t)e3t)u(t)
1 answer
Last reply by: John Lins
Sat Feb 11, 2017 8:06 PM
Post by John Lins on February 11 at 06:25:28 PM
Hello professor vincent. Could you please help me to find this Laplace Transform?
x2 (t)=(1(1t)e3t)u(t)
1 answer
Last reply by: Professor SelhorstJones
Fri Mar 25, 2016 5:24 PM
Post by Ru Chigoba on November 18, 2015
Hi I need help with this problem :
Find the inverse of each relation or function, and then determine if the inverse is a function.
1 f={1,3), (1,1), (1,3), (1,1)} f1=
1 answer
Last reply by: Professor SelhorstJones
Tue Mar 17, 2015 11:24 PM
Post by thelma clarke on March 17, 2015
is there no easier way to solve this it seem very confusing
1 answer
Last reply by: Professor SelhorstJones
Mon Oct 20, 2014 11:27 AM
Post by Saadman Elman on October 18, 2014
It was a great clarification. Thanks,
However, Inverse function is not only has to do with horizontal line test passing but also has to do with vertical test passing. You only stressed on Horizontal test passing. My professor stressed both.
1 answer
Last reply by: Professor SelhorstJones
Mon Jun 16, 2014 9:18 PM
Post by Joshua Jacob on June 15, 2014
Sorry if this is slightly vague but I'm a little but confused on the last example. Could you explain it in other words please?