INSTRUCTORS Carleen Eaton Grant Fraser

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For more information, please see full course syllabus of Algebra 2

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Lecture Comments (8)
 1 answerLast reply by: Dr Carleen EatonThu Mar 27, 2014 6:40 PMPost by HEMAL SINDHVAD on March 9, 2014please this example (âˆœ27)/(âˆš3) I got answer 3^1/4 is it correct or not can you please explain if is wrong 0 answersPost by Prince Dalieh on July 10, 2013Jose Gonalez-Gigato,You are 100% right I think she makes a mistake in example 1 when solving for (f-g)(*). 0 answersPost by julius mogyorossy on June 23, 2013I was experimenting with radicals, I found that what is true in one reality, is not true in the reality, dimension of radicals, I was very surprised to learn this, what does this have to teach me about reality at large, I can't believe I could learn anything more, but maybe I am wrong. 1 answerLast reply by: Norman CervantesSun May 5, 2013 8:20 PMPost by Norman Cervantes on May 5, 201316:25, minor error, 9-6+1 should equal 4 not 2. 1 answerLast reply by: Dr Carleen EatonWed Dec 28, 2011 9:23 PMPost by Jose Gonzalez-Gigato on December 28, 2011In Example I, subtraction should be +4x^4, not -4x^4; the minus signed was incorrectly dropped when writing down the terms within parenthesis.

### Operations on Functions

• The composition of f and g exists only if the range of g is a subset of the domain of f.
• The composition of f and g is almost never equal to the composition in the reverse order.

### Operations on Functions

Let f(x) = 2x3 + 2x2 − 10x + 2 and g(x) = − 4x3 − 2x2 + 4x + 5
Find (f + g)(x),f(x) − g(x)
• Notice that operations on functions is very similar to operations on polynomials.
• When adding, combile the like temrs, when subtracting, distribute the negative sign and make it
• into an addition problem.
• (f + g)(x) = f(x) + g(x)
• = ( 2x3 + 2x2 − 10x + 2 ) + ( − 4x3 − 2x2 + 4x + 5 )
• (f + g)(x) = − 2x2 − 6x + 7
• Now do subtraction, distribute the negative
• (f − g)(x) = f(x) − g(x)
• = ( 2x3 + 2x2 − 10x + 2 ) − ( − 4x3 − 2x2 + 4x + 5 )
• = ( 2x3 + 2x2 − 10x + 2 ) + ( 4x3 + 2x2 − 4x − 5 )
(f − g)(x) = 6x3 + 4x2 − 14x − 3
Let f(x) = − x3 − x2 − x − 1 and g(x) = 4x3 + 6x2 + 4x + 6
Find (f + g)(x),f(x) − g(x)
• Notice that operations on functions is very similar to operations on polynomials.
• When adding, combile the like temrs, when subtracting, distribute the negative sign and make it
• into an addition problem.
• (f + g)(x) = f(x) + g(x)
• = ( − x3 − x2 − x − 1 ) + ( 4x3 + 6x2 + 4x + 6 )
• (f + g)(x) = 3x2 + 5x2 + 3x + 5
• Now do subtraction, distribute the negative
• (f − g)(x) = f(x) − g(x)
• = ( − x3 − x2 − x − 1 ) − ( 4x3 + 6x2 + 4x + 6 )
• = ( − x3 − x2 − x − 1 ) + ( − 4x3 − 6x2 − 4x − 6 )
(f − g)(x) = − 5x3 − 7x2 − 5x − 7
Let g(x) = 2x2 − 2x;h(x) = − 2x3 + 2x
Find (g*h)(x) and ([f/g])(x)
• To multiply, use the foil method, or the distributive method or the box method as covered in previous
• excercises.
• (g*h)(x) = g(x)*h(x) = (2x2 − 2x)( − 2x3 + 2x) = 2x2( − 2x3) + 2x2(2x) − 2x( − 2x3) − 2x(2x)
• Simplify
• (g*h)(x) = − 4x5 + 4x3 + 4x4 − 4x2 = − 4x5 + 4x4 + 4x3 − 4x2
• Divide, factor if necessary. State the constraints in the domain.
• ([f/g])(x) = [f(x)/g(x)] = [(2x2 − 2x)/( − 2x3 + 2x)] = [(2x(x − 2))/(2x( − x2 + 1))] = [((x − 2))/(( − x2 + 1))] = [(x − 2)/( − x2 + 1)]
• To find the restrictions, set the denominator equal to zero.
• − x2 + 1 = 0
• x2 = 1
• x = ±1
([f/g])(x) = [(x − 2)/( − x2 + 1)] except when x = 1,x = − 1
Let g(x) = 2x − 4;h(x) = 4x + 2
Find (g °h)(x),(h °g)(x)
• This is called Function Composition.
• Every where you see an x in g(x) must be replaced by h(x)
• (g °h)(x) = g(h(x)) = 2(h(x) − 4 = 2(4x + 2) − 4
• Simplify
• g(h(x)) = 8x + 4 − 4 = 8x
• To find (h °g)(x),h(g(x)),
• every where you see an x in h(x) must be replaced by g(x)
• (h °g)(x) = h(g(x)) = 4(g(x)) + 2 = 4(2x − 4) + 2
• Simplify
(h °g)(x) = h(g(x)) = 8x − 16 + 2 = 8x − 14
Let g(x) = 2x − 3;h(x) = x3 + 2x2
Find (g °h)(x),(h °g)(x)
• This is called Function Composition.
• Every where you see an x in g(x) must be replaced by h(x)
• (g °h)(x) = g(h(x)) = 2(h(x)) − 3 = 2(x3 + 2x2) − 3
• Simplify
• g(h(x)) = 2x3 + 4x2 − 3
• To find (h °g)(x),h(g(x)),
• every where you see an x in h(x) must be replaced by g(x)
• (h °g)(x) = h(g(x)) = ((g(x))3 + 2(g(x))2 = (2x − 3)3 + 2(2x − 3)
• Simplify
• (h °g)(x) = (2x − 3)(2x − 3)(2x − 3) + 4x2 − 6 = 8x3 − 36x2 + 54x − 27 + 4x2 − 6 =
(h °g)(x) = 8x3 − 32x2 + 54x − 33
Let g(x) = x2 − 2x;h(x) = 3x + 1
Find (g °h)(x),(h °g)(x)
• This is called Function Composition.
• Every where you see an x in g(x) must be replaced by h(x)
• (g °h)(x) = g(h(x)) = (h(x))2 − 2(h(x)) = (3x + 1)2 − 2(3x + 1)
• Simplify
• g(h(x)) = (3x + 1)2 − 2(3x + 1) = 9x2 + 6x + 1 − 6x − 2 = 9x2 − 1
• To find (h °g)(x),h(g(x)),
• every where you see an x in h(x) must be replaced by g(x)
• (h °g)(x) = h(g(x)) = 3(g(x)) + 1 = 3(x2 − 2x) + 1
• Simplify
(h °g)(x) = h(g(x)) = 3(x2 − 2x) + 1 = 3x2 − 6x + 1 =
Let g(x) = x − 1;h(x) = x2 − 1
Find (g °h)(x),(h °g)(x)
• This is called Function Composition.
• Every where you see an x in g(x) must be replaced by h(x)
• (g °h)(x) = g(h(x)) = (h(x)) − 1 = (x2 − 1) − 1 =
• Simplify
• g(h(x)) = x2 − 2
• To find (h °g)(x),h(g(x)),
• every where you see an x in h(x) must be replaced by g(x)
• (h °g)(x) = h(g(x)) = ((g(x))2 − 1 = (x − 1)2 − 1
• Simplify
(h °g)(x) = (x − 1)2 − 1 = x2 − 2x + 1 − 1 = x2 − 2x
Let f(x) = − 4x + 2;g(x) = x2 + 5x
Find (fog)(3)
• You are now evaluating a function composition at a value.
• Evaluate g(3), and the result will become the input to the function f.
• Evaluate f at g(3)
• (fog)(3) = f(g(3)) = f((3)2 + 5(3)) = f(9 + 15) = f(24) = − 4(24) + 2 = − 96 + 2 = − 94
f(g(3)) = − 94
Let f(x) = − 3x − 1;g(x) = x3 − 5x2
Find (fog)(0)
• You are now evaluating a function composition at a value.
• Evaluate g(3), and the result will become the input to the function f.
• Evaluate f at g(3)
• (fog)(3) = f(g(0)) = f((0)3 − 5(0)2) = f(0) = − 3(0) − 1 = − 1
f(g(0)) = − 1
Let f(x) = − 3x − 1;g(x) = x3 − 5x2;h(x) = 2x − 1
Find f(g(h(1)))
• You are now evaluating a function composition at a value.
• Evaluate h(1), and the result will become the input to the function g(x).
• Evaluate g at h(1) and the output will become the input for the function f(x).
• f(g(h(1))) = f(g(2(1) − 1)) = f(g(1)) = f((1)3 − 5(1)2) = f(1 − 5) = f( − 4) = − 3( − 4) − 1 = 12 − 1 = 11
f(g(h(1))) = 11

*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

### Operations on Functions

Lecture Slides are screen-captured 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
• Arithmetic Operations 0:07
• Domain
• Intersection
• Denominator is Zero
• Example: Operations
• Composition of Functions 7:18
• Notation
• Right to Left
• Example: Composition
• Composition is Not Commutative 17:23
• Example: Not Commutative
• Example 1: Function Operations 20:55
• Example 2: Function Operations 24:34
• Example 3: Compositions 27:51
• Example 4: Function Operations 31:09

### Transcription: Operations on Functions

Welcome to Educator.com.0000

Today, we are going to be discussing operations on functions, beginning with arithmetic operations.0002

Recall that two functions can be added, subtracted, multiplied, or divided.0010

The domain of the sum, difference, product, or quotient is the intersections of the domains of the two functions.0016

And remember that intersection, when you are talking about sets, is the areas of the sets that overlap.0024

Often, we will be talking about domains that are all real numbers, and so then there would be complete overlap of those two.0031

But if the sets are slightly different (the domains are slightly different), the domain of these is only0038

the intersection of the domains of the two functions you are performing the operation on.0044

We will talk about when we are working with division, because that is a special case, and there are some additional restrictions on the domain.0049

But let's just go ahead and start out with addition.0059

Let's take, as an example, the two functions f(x) = 2x2 + 5, and a second function g(x) = 3x2 - 4.0062

OK, if you were to add these, then the notation would be as follows: f + g(x).0085

And this can be rewritten to show that it really means f(x) + g(x).0096

So, f + g(x) is just the two functions added together.0104

Looking at what my two functions are: I have 2x2 + 5 as the first function,0110

and I am going to be adding that to my second function, which is 3x2 - 4.0122

And recall that, when you are adding polynomials, all you need to do for addition is remove the parentheses.0128

And that, when we go ahead and add like terms, is going to give us 5x2.0139

And then, combining the constants, 5 - 4 is 1.0144

So, the sum of these two functions is 5x2 + 1.0148

OK, now talking about subtraction: if we were to subtract f - g(x), this can be considered as f(x) - g(x).0155

So, it is the same idea as addition, but you just have to be more careful with the signs.0169

So here, we are going to be taking (2x2 + 5) - (3x2 - 4).0176

Now, when you remove the parentheses, for the first one--this first expression here--this is really positive in front of it;0186

so removing the parentheses is simple--you just take them away; the signs stay the same.0197

To remove the parentheses for this second expression, you have to reverse the signs.0201

So, when I apply the negative sign to 3x2, that is going to give -3x2.0206

Applying the negative sign to the -4 will give me positive 4.0212

And then, I combine these: I get like terms, 2x2 - 3x2, to get -x2; and 4 + 5 (5 + 4) gives me 9.0217

So, that was addition and subtraction of two functions; now, let's talk about multiplication.0231

f times g(x) is the same as f(x) times g(x); this is the same as multiplying two binomials, as we have done in earlier work in this course.0241

So, it is (2x2 + 5) times (3x2 - 4); and since these are binomials,0262

I use the FOIL method: multiplying the first two terms gives me 4 times 3 is 6x4;0271

my outer terms are going to give me -8x2; the inner two terms...5 times 3...that is 15x2.0280

And finally, 5 times -4 gets -20.0291

Simplify by combining like terms: I combine the two x2 terms, and that is going to give me...0295

-8x2 + 15x2, is going to give me positive 7x2; minus 20.0302

OK, so this is multiplication of two functions.0311

Finally, division: division has that additional restriction on the domain, as I mentioned.0317

Let's look at dividing here: f divided by g(x), which is actually the same as f(x) over g(x).0324

f(x) is 2x2 + 5...over 3x2 - 4.0338

Now, when you are handling division, you need to think about excluded values in the denominator.0347

As always, if the denominator is 0, that would result in an expression that is undefined.0354

You cannot divide by 0, so we have to think about situations where this expression,0360

3x2 - 4, would equal 0, and then exclude those from the domain.0366

So, I am going to go ahead and solve for x, and see what values of x would make this expression 0.0371

I can do that by adding 4 to both sides, dividing both sides by 3, and then taking the square root of both sides.0378

This tells me that the excluded values, when performing this operation (performing division and finding the quotient) are x = √ or -√4/3.0394

So, x cannot equal ±√4/3, because if it does, the result will be a 0 in the denominator and a value that is undefined.0419

So, this is a situation that applies only to division.0430

Composition of functions: suppose that f and g are functions, and the range of g, the second function, is a subset of the domain of f, the first function.0438

I will talk a little bit more about that in a second.0453

But right now, let's look at what the notation is, and what it is really saying when we talk about composition of functions.0456

The composition of f and g is defined by...this symbol, this open circle, means "composed with."0465

It is not a multiplication symbol: the open circle would be read as "f composed with g of x."0476

And that is going to result in a composite function.0483

And another way to write this, and to think about it, is that what this is saying is f(g(x)).0486

Something to keep in mind as we work with composition of functions is that you actually move from right to left when you are working with these.0493

So, when I consider a composite function (and it could actually be composed of three functions or four functions),0502

I am going to start out by working my way from the right, seeing if I am given a value here--0509

instead of g(x), it could be g(2)--and then working with that, determining what g(x), and then plugging that value into my f function.0515

OK, so let's first use an example: if I have two functions--one is f(x) = x2 - 2x + 1,0528

and another one is g(x) = 4x - 5--I might be asked to find the composition of those two functions.0538

And that would be...I could be asked to find f composed with g, or g composed with f; but I am going to work with f composed with g.0548

Again, I am rewriting that as...this is saying it is f of g of x.0564

So, I am going to start here with this function on the right.0571

And I am going to say, "OK, this is really saying f(g(x))."0574

Well, I look here, and g(x) is 4x - 5.0577

Previously, we have talked about evaluating a function for an algebraic expression.0585

We first started out by evaluating functions for numbers, like f(3): I would just substitute all of the x's with 3's.0591

I also talked about finding f for an algebraic expression; and that is really what we end up doing here.0601

So, if I am trying to find f(4x - 5), wherever I see an x, I just replace it with this expression, 4x - 5...0606

which is going to be...here I have x2, so I am going to replace that x with 4x - 5 and square it.0615

I have a second x right here; I have -2 times whatever is in here, since that is replacing x.0625

And then, I am adding 1; so right here, where there is an x, simply substitute 4x - 5.0633

And then, we are going to figure what this would become equal to.0641

And this is going to give me...if you recall, if I square a binomial, if I form a perfect square trinomial from that...0645

it ends up being this first term squared (4x, squared), and then 2 times the product of these two terms;0658

so this is going to give me 4x, times -5, plus the square of this last term.0669

You could always write this out as (4x - 5) (4x - 5) and use FOIL if you didn't remember that rule.0678

OK, then over here, I am going to multiply everything inside the parentheses by -2 to get -8x + 10 + 1.0685

OK, so this is going to give me...I am actually rewriting this like that, because I also have to square the 4.0698

So, this is going to give me (4x)2, so that will actually be 16x2,0709

and then 2 times 4x is 8x, times -5 is going to give me 8x times -5...-40x.0717

And then here, I have -5 times -5, so that is + 25, minus 8x plus 10 plus 1.0734

Simplifying, I still have 15x2; combining -40x and -8x is going to be -48x.0743

25 + 10 + 1 is 35...36.0753

f composed with g: I went about finding this by first replacing this with g(x), which is 4x - 5,0762

and then finding f of that by replacing all of the x's here with 4x - 5.0769

And then, I squared this binomial, took 2 times the binomial here, and then added 1.0776

And this is the result: f composed with g of x equals this.0783

Now, talking a little bit more about domain and range: let's say that we were asked to find f composed with g(2).0790

Well, I know what f composed with g(x) is, and I could just plug the 2 in here.0802

But let's say we hadn't done that work--that we were just given these two, and then told, "OK, find f composed with g(2)."0808

So, I am just going to start there and think about how this works.0814

Well, I start at the right; I am going to rewrite this here, first, as f(g(x)).0818

In this case, I said that x is going to be 2; so I am going to put a 2 there.0828

So, I am going to find g(2), and then I am going to find f of that value.0833

So, this is going to equal f of...well, g(x) is 4x - 5, but I am asked to find g(2).0838

So remember that g(2) is going to be 4 times 2, minus 5.0847

I am going to put that in here: 4 times 2, minus 5.0854

OK, that equals f of 8 - 5, which is 3, so I am being asked to find f(3).0863

Now, let's think about this for a second: here, my result, 3, is an element of the range of g.0874

My input value for g--my domain value--is 2; I evaluated that for this function g, and I came up with 3.0894

g(2) is 3, so 3 is part of the range.0903

Now, I am using that as my input value for the function f.0908

So, 3 is an element of the range of g, and an element of the domain of f.0914

The range of g is a subset of the domain of f; that is what this is saying up here.0928

So, I evaluate g for a particular value--just the general case x, or a certain number.0934

I find what that is: that is a part of this range; it is also an element of the domain of f.0942

The entire set of values--the entire range for g--is a subset of the domain of f.0950

So, let's go ahead and finish this out: I found g(2) to be 3; now I am going to evaluate f for 3.0963

Well, f is x2 - 2x + 1, so wherever I see an x, I am simply going to substitute in a 3.0972

That is going to give me 9 - (6 + 1); that is 9 - 7, so that is 2.0982

So, f composed with g(2)...actually, that is 4--a correction right there: f composed with g(2) is 4.0992

So again, starting out, we are going right to left, finding g(2), substituting what I came up with,1013

which is 3, right in here, finding f(3), and evaluating for that.1024

And I find that f composed with g(2) is 4.1029

Up here, I just talked about the general case, f composed with g(x), and I found this.1034

OK, so that is an introduction to composition of functions.1040

And one important thing to keep in mind is that composition is not commutative.1043

What we mean by that is that, in general, f composed with g does not equal g composed with f.1048

We will see, later on, cases where we will talk about this more, later;1055

but in general, you can't just flip these two around and assume that the opposite case is equal, because it is actually usually not.1061

Illustrating that with an example: let's let f(x) equals x2 + 1, and g(x) equal 2x - 3.1071

Now, I am going to find both f composed with g and g composed with f, and then compare what I get.1091

So, f composed with g(x) = f(g(x)); starting from the right, let's put g(x) in here, which is 2x - 3.1096

Now, I am evaluating f for this expression, 2x - 3.1115

And f is x2 + 1, so I am going to substitute 2x - 3 for this x, and square it, and then add 1.1120

Squaring this would give me 2x squared, plus 2 times 2x times -3, plus -3 squared, plus 1.1133

OK, that gives me 4x2; 2 times 2 is 4, times -3 is -12x; plus -32, which is 9, plus 1.1150

Simplify to 4x2 - 12x + 10.1166

f composed with g(x) is equal to this; now, let's try g composed with f(x), which is equal to g(f(x)).1173

f(x) is x2 + 1; I am going to put that in here.1190

Now, I am going to evaluate g for x2 + 1 by substituting x2 + 1 everywhere there is an x in this function.1195

That is going to give me 2 times x2 + 1 - 3.1206

2 times x2 is 2x2; plus 2 times 1 is 2; minus 3 gives me 2x2 - 1.1217

And this is g composed with f(x).1227

As you can see, f composed with g(x) is not the same as g composed with f.1232

So, I can't make that assumption, that these two are the same.1245

In fact, if anything, it is likely that they are not the same; composition is not commutative.1248

The first example: in this example, we are given two functions, f(x) and g(x), and asked to find their sum and their difference.1257

Starting out with addition: recall that f + g(x) = f(x) + g(x).1265

So, I am simply going to add the two functions; f(x) is 3x3 - 2x2 - x - 1,1274

plus -4x4 + 2x3 - 3x - 4.1286

OK, since this is addition, I can simply remove the parentheses without worrying about having a problem with the signs.1294

The signs remain the same, so I just continue to retain the signs in here, since this is addition.1302

I can just take the parentheses away.1313

Let's add this, and at the same time put it in descending order to help keep track of everything.1320

The largest power I have is 4; so I have -4x4, and there is no like term that I can combine that with.1326

For terms that are cubed, I have 3x3 and 2x3 to give me 5x3.1335

For terms that are squared, I only have one term, and that is -2x2.1342

For x's, I have -x and -3x to give -4x; and for constants, -1 and -4 are -5.1348

So, f + g(x) equals this expression.1360

Subtraction: f - g(x) = f(x) - g(x); now here, I do need to be careful with the signs.1370

So, I am rewriting: 3x3 - 2x2 - x - 1, minus g(x), which is 4x4 + 2x3 - 3x - 4.1384

OK, removing the parentheses: since this is a positive in front of this, it remains 3x3...1398

Here, I need to apply the negative sign to each term inside the parentheses; and I can do that by reversing the sign.1413

So, this becomes -4x4; -2x3; a negative and a negative--that gives me a positive 3x; and a negative and a negative...plus 4.1419

Combining like terms, I get -4x4; here, I have 3x3 and -2x3, leaving + x3.1432

I only have one x2 term, so I leave that alone; that is -2x2.1447

For x's, -x + 3x is going to leave me with just 2x, and then the constants: -1 + 4 gives me _ 3.1453

So, f - g(x) is given by this expression right here.1466

Example 2: this time, we are given two functions, f(x) and g(x), and told to find the product and the quotient of these functions.1474

OK, beginning with multiplication: the product f times g(x) equals f(x) times g(x),1485

which equals (4x2 - 7) times (3x2 + 9x).1499

So, the first terms--multiplying those, that is 12x4; the outer terms give me 4 times 9; that is 36x2 times x, so that is 36x3.1513

The inner two terms--that is -21x2; and then -7 times 9 is going to give me -63; and I have an x here.1528

OK, and looking at this, I can't simplify this any further, because I don't have any like terms.1539

So, I am just going to leave that alone; and this gives me f times g of x.1547

OK, the second task is to find the quotient, f divided by g of x, which equals f(x) divided by g(x).1553

which equals (4x2 - 7) divided by (3x2 + 9x).1567

Now, recall: when working with division of functions, we need to find the excluded values.1578

And the excluded values are those values of x, those numbers of the domain, that will make the denominator 0,1582

because if this denominator is 0, I will have an undefined expression.1590

So, any value of x that makes 3x2 + 9x equal 0 is excluded from the domain.1594

Handling this by factoring: I can see that I have a greatest common factor of 3x.1604

I have a 3 here and a 3 in here, and I have an x here and an x in here that I can pull out.1611

This will leave behind an x and a 3.1615

Using the zero product property, this tells me that 0 = 3x, or it could be that x + 3 = 0.1620

If either of those is true, this product becomes 0.1633

OK, so this will end up giving me (let's rewrite this as) 3x = 0, so divide both sides by 3: x = 0--that is an excluded value.1638

And x = -3--these are excluded values, meaning this function is not defined for domain values of x = 0 and x = -3.1652

Those values are excluded from the domain.1667

Example 3: now, we are going to work with composition of functions.1671

Given f(x) and g(x), we are asked to find f composed with g(x) and g composed with f(x), starting with this one.1677

OK, so recall that f composed with g(x) is the same as saying f(g(x)).1686

So, I am going to start from the right and go to the left.1698

First, I am going to figure out what g(x) is; and I have that given right here as 4x2.1702

I am asked to find f of that; I am going to evaluate f for 4x2.1710

So, wherever I see an x in f(x), in this function, I will substitute 4x2.1715

So, this is going to give me 2 times x; and in this case, x is 4x2; minus 3.1721

What I have here is rewritten here, just with 4x2 replacing the x.1729

OK, that is going to give me 8x2 - 3; so, f composed with g of x equals 8x2 - 3.1737

Now, g composed with f(x) equals g(f(x)), starting from the right: what do I have in here?1753

I have f(x), which is 2x - 3; so this equals g(2x - 3).1766

This means I need to evaluate the function g when x is defined as 2x - 3.1774

So, I have an x here; it is 4 times x2, so 4(2x - 3)2, just replacing this with this.1785

OK, squaring 2x - 3 and multiplying that times 4 is going to give me (2x)2 + 2 times 2x times -3 + (-3)2.1795

That is going to give me 4 times...this is 2 times 2...that is going to give me 4x2;1816

2 times 2x is 4x, times -3; that is -12x; (-3)2 is 9.1826

Multiplying each of these by 4 gives me 16x2; 4 times -12 is -48x; 4 times 9 is 36.1833

And this also illustrates what we talked about before, that composition is not commutative.1846

f composed with g of x is not equal to g composed with f of x.1854

It may be, but it is not necessarily true; you cannot assume that it is true.1864

This time, we are going to be working with three functions: f(x), g(x), and h(x).1870

And we handle these the same way as we did composition of functions, when we were only working with two functions.1876

And we are going to work from right to left.1884

Here, instead of just asking for f or g or h of x, they are asking me for f(g(h(the particular value -2))).1887

OK, so I am going to start out by looking for h(-2).1897

Let's go ahead and find that value; I want to find f(g(h(-2))).1906

So, what is h(-2)? Well, h is 2x2 + 3; therefore, h(-2)...I would have to substitute in 2...I put a -2 here and square it...+ 3.1913

So, this is f(g(2 times -2 squared, plus 3)), which equals f(g(...-2 squared is 4, plus 3)).1926

So, this gives me 8 + 3, which is f(g(11)); so starting from the right, I evaluated h for -2; that gave me 11.1952

Now, this is a member of the range of h, and it is also an element of the domain of g, because I am plugging it in here as an input value.1971

So now, I need to find g(11): well, g(x) equals 4x - 2; so substituting 4 times 11 - 2 is what I am going to be doing here.1981

f(4 times 11 minus 2) equals 44 - 2, equals 42; so, g(11) = 42.2002

Finally, I am just left with f(42); so I need to evaluate f(x) when x is 42.2020

So, I simply substitute in, and that is going to give me 6 times 42 (which is 252).2031

So, the result from this composition of functions is 252, from evaluating h for -2; finding that result;2043

evaluating g for that result (which was h(-2), which was 11); evaluating g for 11;2052

finding that that was 42; and then evaluating f for that value.2060

That concludes this session of Educator.com; thanks for visiting!2066