### 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

^{3}+ 2x

^{2}− 10x + 2 and g(x) = − 4x

^{3}− 2x

^{2}+ 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)
- = ( 2x
^{3}+ 2x^{2}− 10x + 2 ) + ( − 4x^{3}− 2x^{2}+ 4x + 5 ) - (f + g)(x) = − 2x
^{2}− 6x + 7 - Now do subtraction, distribute the negative
- (f − g)(x) = f(x) − g(x)
- = ( 2x
^{3}+ 2x^{2}− 10x + 2 ) − ( − 4x^{3}− 2x^{2}+ 4x + 5 ) - = ( 2x
^{3}+ 2x^{2}− 10x + 2 ) + ( 4x^{3}+ 2x^{2}− 4x − 5 )

^{3}+ 4x

^{2}− 14x − 3

^{3}− x

^{2}− x − 1 and g(x) = 4x

^{3}+ 6x

^{2}+ 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)
- = ( − x
^{3}− x^{2}− x − 1 ) + ( 4x^{3}+ 6x^{2}+ 4x + 6 ) - (f + g)(x) = 3x
^{2}+ 5x^{2}+ 3x + 5 - Now do subtraction, distribute the negative
- (f − g)(x) = f(x) − g(x)
- = ( − x
^{3}− x^{2}− x − 1 ) − ( 4x^{3}+ 6x^{2}+ 4x + 6 ) - = ( − x
^{3}− x^{2}− x − 1 ) + ( − 4x^{3}− 6x^{2}− 4x − 6 )

^{3}− 7x

^{2}− 5x − 7

^{2}− 2x;h(x) = − 2x

^{3}+ 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) = (2x
^{2}− 2x)( − 2x^{3}+ 2x) = 2x^{2}( − 2x^{3}) + 2x^{2}(2x) − 2x( − 2x^{3}) − 2x(2x) - Simplify
- (g*h)(x) = − 4x
^{5}+ 4x^{3}+ 4x^{4}− 4x^{2}= − 4x^{5}+ 4x^{4}+ 4x^{3}− 4x^{2} - Divide, factor if necessary. State the constraints in the domain.
- ([f/g])(x) = [f(x)/g(x)] = [(2x
^{2}− 2x)/( − 2x^{3}+ 2x)] = [(2x(x − 2))/(2x( − x^{2}+ 1))] = [((x − 2))/(( − x^{2}+ 1))] = [(x − 2)/( − x^{2}+ 1)] - To find the restrictions, set the denominator equal to zero.
- − x
^{2}+ 1 = 0 - x
^{2}= 1 - x = ±1

^{2}+ 1)] except when x = 1,x = − 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)) = 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

^{3}+ 2x

^{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)) − 3 = 2(x
^{3}+ 2x^{2}) − 3 - Simplify
- g(h(x)) = 2x
^{3}+ 4x^{2}− 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) + 4x
^{2}− 6 = 8x^{3}− 36x^{2}+ 54x − 27 + 4x^{2}− 6 =

^{3}− 32x

^{2}+ 54x − 33

^{2}− 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) = 9x^{2}+ 6x + 1 − 6x − 2 = 9x^{2}− 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(x
^{2}− 2x) + 1 - Simplify

^{2}− 2x) + 1 = 3x

^{2}− 6x + 1 =

^{2}− 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 = (x
^{2}− 1) − 1 = - Simplify
- g(h(x)) = x
^{2}− 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

^{2}− 1 = x

^{2}− 2x + 1 − 1 = x

^{2}− 2x

^{2}+ 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

^{3}− 5x

^{2}

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

^{3}− 5x

^{2};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

*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

### Algebra 2

### 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 only*0038

*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) = 2x ^{2} + 5, and a second function g(x) = 3x^{2} - 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 2x ^{2} + 5 as the first function,*0110

*and I am going to be adding that to my second function, which is 3x ^{2} - 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 5x ^{2}.*0139

*And then, combining the constants, 5 - 4 is 1.*0144

*So, the sum of these two functions is 5x ^{2} + 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 (2x ^{2} + 5) - (3x^{2} - 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 3x ^{2}, that is going to give -3x^{2}.*0206

*Applying the negative sign to the -4 will give me positive 4.*0212

*And then, I combine these: I get like terms, 2x ^{2} - 3x^{2}, to get -x^{2}; 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 (2x ^{2} + 5) times (3x^{2} - 4); and since these are binomials,*0262

*I use the FOIL method: multiplying the first two terms gives me 4 times 3 is 6x ^{4};*0271

*my outer terms are going to give me -8x ^{2}; the inner two terms...5 times 3...that is 15x^{2}.*0280

*And finally, 5 times -4 gets -20.*0291

*Simplify by combining like terms: I combine the two x ^{2} terms, and that is going to give me...*0295

*-8x ^{2} + 15x^{2}, is going to give me positive 7x^{2}; 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 2x ^{2} + 5...over 3x^{2} - 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

*3x ^{2} - 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) = x ^{2} - 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 x ^{2}, 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 16x^{2},*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 15x ^{2}; 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 x ^{2} - 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 x ^{2} + 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 x ^{2} + 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 4x ^{2}; 2 times 2 is 4, times -3 is -12x; plus -3^{2}, which is 9, plus 1.*1150

*Simplify to 4x ^{2} - 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 x ^{2} + 1; I am going to put that in here.*1190

*Now, I am going to evaluate g for x ^{2} + 1 by substituting x^{2} + 1 everywhere there is an x in this function.*1195

*That is going to give me 2 times x ^{2} + 1 - 3.*1206

*2 times x ^{2} is 2x^{2}; plus 2 times 1 is 2; minus 3 gives me 2x^{2} - 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 3x ^{3} - 2x^{2} - x - 1,*1274

*plus -4x ^{4} + 2x^{3} - 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 -4x ^{4}, and there is no like term that I can combine that with.*1326

*For terms that are cubed, I have 3x ^{3} and 2x^{3} to give me 5x^{3}.*1335

*For terms that are squared, I only have one term, and that is -2x ^{2}.*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: 3x ^{3} - 2x^{2} - x - 1, minus g(x), which is 4x^{4} + 2x^{3} - 3x - 4.*1384

*OK, removing the parentheses: since this is a positive in front of this, it remains 3x ^{3}...*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 -4x ^{4}; -2x^{3}; 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 -4x ^{4}; here, I have 3x^{3} and -2x^{3}, leaving + x^{3}.*1432

*I only have one x ^{2} term, so I leave that alone; that is -2x^{2}.*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 (4x ^{2} - 7) times (3x^{2} + 9x).*1499

*So, the first terms--multiplying those, that is 12x ^{4}; the outer terms give me 4 times 9; that is 36x^{2} times x, so that is 36x^{3}.*1513

*The inner two terms--that is -21x ^{2}; 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 (4x ^{2} - 7) divided by (3x^{2} + 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 3x ^{2} + 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 4x ^{2}.*1702

*I am asked to find f of that; I am going to evaluate f for 4x ^{2}.*1710

*So, wherever I see an x in f(x), in this function, I will substitute 4x ^{2}.*1715

*So, this is going to give me 2 times x; and in this case, x is 4x ^{2}; minus 3.*1721

*What I have here is rewritten here, just with 4x ^{2} replacing the x.*1729

*OK, that is going to give me 8x ^{2} - 3; so, f composed with g of x equals 8x^{2} - 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 x ^{2}, 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 4x ^{2};*1816

*2 times 2x is 4x, times -3; that is -12x; (-3) ^{2} is 9.*1826

*Multiplying each of these by 4 gives me 16x ^{2}; 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 2x ^{2} + 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

1 answer

Last reply by: Dr Carleen Eaton

Thu Mar 27, 2014 6:40 PM

Post by HEMAL SINDHVAD on March 9, 2014

please this example (Ã¢Ë†Å“27)/(Ã¢Ë†Å¡3)

I got answer 3^1/4 is it correct or not can you please explain if is wrong

0 answers

Post by Prince Dalieh on July 10, 2013

Jose Gonalez-Gigato,You are 100% right I think she makes a mistake in example 1 when solving for (f-g)(*).

0 answers

Post by julius mogyorossy on June 23, 2013

I 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 answer

Last reply by: Norman Cervantes

Sun May 5, 2013 8:20 PM

Post by Norman Cervantes on May 5, 2013

16:25, minor error, 9-6+1 should equal 4 not 2.

1 answer

Last reply by: Dr Carleen Eaton

Wed Dec 28, 2011 9:23 PM

Post by Jose Gonzalez-Gigato on December 28, 2011

In Example I, subtraction should be +4x^4, not -4x^4; the minus signed was incorrectly dropped when writing down the terms within parenthesis.