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(a) If $ g(x) = 2x + 1 $ and $ h(x) = 4x^2 + 4x + 7 $, find a function $ f $ such that $ f \circ g = h $. (Think about what operations you would have to perform on the formula for $ g $ to end up with the formula for $ h $.)(b) If $ f(x) = 3x + 5 $ and $ h(x) = 3x^2 + 3x + 2 $, find a function $ g $ such that $ f \circ g = h $.

(a) By examining the variable terms in $g$ and $h$, we deduce that we must square $g$ to get the terms $4 x^{2}$ and $4 x$ in $h$. If we let \[ f(x)=x^{2}+c, \text { then }(f \circ g)(x)=f(g(x))=f(2 x+1)=(2 x+1)^{2}+c=4 x^{2}+4 x+(1+c) . \text { since }\] $h(x)=4 x^{2}+4 x+7,$ we must have $1+c=7,$ So $c=6$ and $f(x)=x^{2}+6$.(b) We need a function $g$ so that $f(g(x))=3(g(x))+5=h(x)$. But $h(x)=3 x^{2}+3 x+2=3\left(x^{2}+x\right)+2=3\left(x^{2}+x-1\right)+5,$ so we see that $g(x)=x^{2}+x-1$.

04:27

Jeffrey P.

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 3

New Functions from Old Functions

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Johns Hopkins University

University of Michigan - Ann Arbor

Idaho State University

Boston College

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So here we have function h of X, and we know that h of X is f f G G is the inside function and we know it. But we don't know the outside function. That's what we're going to find. We're going to find F. So let's find a way to rewrite H. And when you see four x squared here and you see two X here, hopefully you realize that in some way we must have squared G. So let's figure out what G squared is. G squared would be two x plus one quantity squared, and that would be four x squared plus four x plus one. So let's think of age as four X squared plus four x plus one plus six. We have to get that seven in there, so we need one in six to give us seven. And now we can rewrite that perfect try no meal square as two x plus one quantity squared and then we have the plus six. So if we think of age in this way, we can see the inside function. G is two x plus one on the outside function. Must be. Then what? We get when we square that and then add six. So f of X is X squared plus six for part B. We have a similar situation where we have a function h of X, and now we know the outside function f but we don't know the inside function G. That's what we have to figure out. So what we want to do is work on H and get it to look more like this three times something plus five. So if we change h a little bit by factoring out of three, we have three times X squared plus X plus two. Well, we need that to be plus five on the end, not plus two. So we need to add another three. But how can we just add three? We can't just add three unless we also subtract three. So what I'm going to dio is incorporate this minus three into the inside of the parentheses in some way. So if I write it as three times a quantity X squared plus X minus one and then the two plus three on the outside is going to be plus five. Here we have our minus three on the inside compensating for are extra plus three on the outside. So we have three times a quantity X squared plus X minus one plus five that is H of X, rewritten in a different form. Now that matches with our f of X three times something plus five. So that's something on the inside that must be RG so g of X is X squared plus X minus one.

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