WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today we will be covering inverse functions and relations.
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And this is a topic (inverse relations) that we introduced a little bit in Algebra I.
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But we will be going into much greater depth today.
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First, reviewing the definition of the inverse of a relation: recall that a **relation** is a set of ordered pairs.
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The inverse relation of relation R (and the notation is this, R^-1, or if you had a different letter,
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when we are talking about functions, then you would put that letter, and then -1,
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to express that you are talking about the inverse of a relation) is the set of ordered pairs
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in which the coordinates of each ordered pair of R are reversed.
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So, let's think about what this means.
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First of all, a relation is a set of ordered pairs; and that would be something like {(2,4),(3,-7),(2,6),(8,-4)}.
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And this is a relation, because there is a correspondence between the first element and the second element.
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You could write these out as a table; and each of the first elements has a correspondence with a member of the second set.
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So, this is the relation; the inverse of R, R^-1, would be when the first and second coordinates are switched.
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What we are doing, then, is interchanging the domain and the range.
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So, 2, 3, 2, and 8 are members of the domain; and 4, -7, 6, -4 are members of the range; now it becomes the opposite.
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Now, 4, -7, 6, -4 are the domain, and these other values are the range.
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I also want to point something out: let's write R as a table, and I have my x-values, 2, 3, 2, 8, and my y-values, 4, -7, 6, -4.
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This is a relation, because there is a correspondence between members of the x (the domain), and members of the range (the y-values).
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2 corresponds to 4, 3 to -7, and so on.
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So, this is a relation; however, it is not a function; and that is because you see that 2 is assigned to two values of the range.
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And in a function, each member of the domain is assigned to only one member of the range.
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So, this is a relation--yes; function--no; it is not a function.
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Functions are relations, but not all relations are functions; functions are a subset of relations.
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Let's talk more about functions right now.
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The inverse function (f^-1(x)) is a special case of an inverse relation; the inverse of a function f(x) is a special case.
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And, as I talked about, there are some restrictions on functions.
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Functions must meet the criteria that each element of the domain is assigned to only one member of the range.
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f^-1(b) = a if and only if f(a) = b.
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The domain and range of the inverse function are the range and domain of the function.
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Just as I talked about with relations, the domain and range are flipped; they are switched around
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when you go from the function to the inverse, or from the inverse to the function.
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And let's think about what this is saying right here: f^-1(b) = a if and only if f(a) = b.
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Let's look at what that would mean in a general case.
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If I have some function, and then I said part of this function is the ordered pair (3,4),
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if it has an inverse, then that means that the values are going to be reversed.
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If I looked for f(3), I would get 4 out; if I went ahead and put 4 into the inverse, my result would be 3; domain and range are switched around.
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Taking a more specific example, if f(x) = 2x - 1, and f^-1(x) is (x + 1)/2, let's say I was asked to find f(6).
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Then, I would substitute that 6 in here to get 11.
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My expectation would be that f^-1(11) would be 6, because, if f(a) = b,
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and here we are letting a equal 6 and b equal 11, then f^-1(b), f^-1(11), should equal 6.
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So, let's check out if that holds up--let's see if it does.
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I am asked to find f^-1(11), which would be 11 + 1, over 2, equals 12/2, equals 6.
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So, f^-1(11) is 6, and f(6) is 11; so, this holds up and supports the fact that these two are inverse functions of each other.
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There is a certain set of rules that you can follow to actually construct an inverse function.
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So, if f(x) is given, f^-1 can be constructed by following this procedure.
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First, write y instead of f(x): change the f(x) notation to y.
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Then, interchange x and y: wherever there is an x, make it a y; wherever there is a y, change that to an x.
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The hardest step is this one: you need to then solve for y, which, depending on the function, can be somewhat complicated.
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Then, write f^-1(x) for y: once you have solved for y (you have isolated y on the left side of the equation),
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you rewrite that as f^-1, because that is what you are looking for.
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For example, f(x) = 2x + 7; if you wanted to find f^-1(x), step 1 is to write y instead of f(x).
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1, 2, 3, 4 steps: the first step is to write y instead of f(x).
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The second step is to interchange x and y: I am going to change that to an x; I am going to change that to a y.
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The third step is to solve for y: I am going to subtract 7 from both sides--I am trying to isolate y.
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Now, I need to divide both sides by 2.
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I have y isolated, but I am just going to rewrite this in a more standard form, with the isolated variable on the left side of the equation.
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So, I have y = (x - 7)/2, and now I am going to write it as f^-1(x), instead of using a y here.
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I have found the inverse of f(x) by following these four steps.
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One more example of constructing an inverse function: given f(x) = 5x - 3, we are going to find f^-1(x).
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Step 1: Change f(x) to y--let's rewrite this over here and follow each step down.
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So, step 1--I am going to change this to y = 5x - 3.
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Step 2: Interchange x and y--I am going to make this an x; I am going to make this a y.
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Then, solve for y.
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This was step 1; step 2; solving for y--I am going to add 3 to both sides;
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then I need to divide both sides by 5; so I get (x + 3)/5 = y; and I am just going to rewrite that with the y on the left, in standard form here.
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OK, next, replace y with f^-1(x)--just change the notation.
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This is going to be f^-1(x) = (x + 3)/5.
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So, I found the inverse of this function by using this four-step procedure.
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Inverses and compositions: in the last lesson, we talked about composition functions.
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And now, we are going to discuss how this relates to inverses, and how it can help you to determine if two functions are inverses of each other.
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The functions f and g are inverses if and only if f composed with g of x equals x, and g composed with f of x equals x, for all x.
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So, if I put a number in here (2), and I evaluate it for the composite function, I will get that same value back.
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If I use 2, I will get 2 back; if I use 50, I will get 50 back; and the same when I look at g composed with f.
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If I evaluate that for a particular value, 7, I will get 7 back.
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Let's go ahead and take an example to illustrate this.
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Given f(x) = 6x - 1, and g(x) = (x + 1)/6, are they inverses?
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I can use this to evaluate whether or not two functions are inverses of each other.
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And if they are inverses, then when I take f composed with g of x, I should get x back.
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Let's look at what f composed with g of x is: recall that is the same as saying f(g(x)).
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Replace this with g(x): well, g(x) is (x + 1)/6; f of this can be found by looking at the function and replacing x with this algebraic expression, (x + 1)/6.
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So, f((x + 1)/6) is 6 times (x + 1)/6, minus 1.
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And here, these 6's end up cancelling out; so that leaves me with x + 1 - 1 = x.
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So, f composed with g of x does equal x; that is half of what I have to figure out.
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Now, does g composed with f of x equal x?
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Well, g composed with f of x equals g(f(x)); well, that is g(6x - 1), so I need to evaluate g for this algebraic expression.
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Well, g is (x + 1)/6, so I need to substitute 6x - 1 right here, and then I need to add 1 to that value and divide it by 6.
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I am going to remove these parentheses to get (6x - 1 + 1)/6.
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The 1's cancel each other out to leave 6x/6; the 6's cancel to get x; so, g composed with f of x equals x.
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And f composed with g of x is x, so are they inverses? Yes, f(x) and g(x) are inverses of each other.
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So, this fact right here allows us to determine if two functions are inverses of each other or not.
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OK, the first example: find the inverses of the relation (I am going to call this relation R, and I am asked to find the inverse of R).
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Remember that, in the inverse, I flip around the two values; these domain values and the range values are going to be reversed.
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That is going to give me {(3,1),(4,2),(7,3),(8,1),(9,4),(6,3)}; and then, I am just double-checking that they are all correctly reversed; and they are.
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This is the inverse of the relation R.
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Next, find the inverse of g(x) = 2x + 4.
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All right, the first step is to change g(x) to y; that is going to give me y = 2x + 4.
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Second, interchange x and y: OK, this is going to become x; this will become y.
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The third step: solve for y--I am going to subtract both sides by 4: x - 4 = 2y.
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Next, I am going to divide both sides by 2: (x - 4)/2 = y.
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I have isolated y; I am just going to rewrite this with y on the left.
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And then, I am going to replace y, or change y to the notation g^-1(x).
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I found that the inverse of this function is (x - 4)/2 by following these four steps.
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Example 3: Find the inverse of h(x) = 2/3x - 4.
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Again, here it is h(x), so change h(x) to y; that is going to give me y = 2/3x - 4.
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The second step: interchange x and y--this is going to give me x = 2/3y - 4.
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Next, solve for y: I need to add 4 to both sides, and then, after than, multiply both sides by 3/2.
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And that is going to give me 3/2(x + 4) = y; multiplying this out, that is going to give me 3/2x + 3/2(4) = y.
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3/2x...and this is 3 times 4 is 12, divided by 2; that becomes 6.
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Finally, I am going to change y to h^-1(x).
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I am going to do that at the same time as I am going to go ahead and put the y on the left side, and this becomes h^-1(x) = 3/2x + 6.
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So, the inverse of the function given is h^-1(x) = 3/2x + 6--again, I did this by following the four-step procedure.
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Given f(x) and g(x), we are asked to determine if f and g are inverses of each other.
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So, are f and g inverses of each other?
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Now, recall that two functions are inverses of each other if and only if this scenario holds up.
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If f composed with g of x equals x, and g composed with f of x equals x, then the functions are inverses.
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OK, so let's go ahead and find f composed with g of x, which equals f(g(x)).
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OK, f(g(x))--well, that is f of this expression--f(1/5(x + 8)).
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Well, f(x) is 5x - 8, so f of this expression would be 5; then, replacing x with this, that is 1/5(x + 8) - 8.
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OK, let's get rid of these parentheses and do some multiplying.
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This is 5 times 1/5(x + 8) - 8; these 5's cancel, and I am just left with a 1 here; so that is just (x + 8) - 8 = x + 8 - 8 = x.
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So, we have our first part: f composed with g of x is x; now, let's see if it works out for g composed with f.
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Well, recall that that is going to be the same as g(f(x)).
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Up here, f(x) is 5x - 8; so g(5x - 8) is what we are looking for; I need to evaluate g for this expression, this function, when x is 5x - 8.
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So, that is 1/5...of (5x - 8); and then I am going to add 8 to that.
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So, here I have 1/5...now let's get rid of some of these parentheses inside to simplify...5x - 8 + 8, getting rid of these...
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That is 1/5, and then I have a negative 8 and a positive 8, so those become 0; so it is just 1/5 times 5x.
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The 5's cancel out, so that equals x; so g composed with f of x is also x.
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So, are these inverses of each other? Yes, f(x) and g(x) are inverses of each other, because these two hold true.
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f composed with g of x is x, and g composed with f of x is also x.
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That concludes this lesson of Educator.com; thanks for visiting!