The function is defined by The function is defined by . For what value of does reach its maximum?

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SAT Nonlinear Functions Question 63 | Free SAT Prep | Aniko

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Question 63·Hard·Nonlinear Functions

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The function $f$ is defined by

f(x)=2^{x-1}+2^{1-x}.

The function $g$ is defined by $g(x)=5-3\,f(2x+4)$ . For what value of $x$ does $g(x)$ reach its maximum?

For composite functions of the form $g(x)=a+b\,f(cx+d)$ , first notice whether $b$ is positive or negative: if $b<0$ , maximizing $g$ is the same as minimizing $f$ at the transformed input. Then reduce the problem by renaming the inner linear expression (like $t=2x+4$ ) and analyze $f(t)$ on its own. Look for algebraic simplifications or substitutions (such as rewriting $2^{1-t}$ as $1/2^{t-1}$ and letting $u=2^{t-1}$ ) that turn the expression into a simple pattern like $u+1/u$ , which you can minimize using basic inequalities or symmetry, and finally convert back to $x$ using the linear relation.

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Look at the structure of $g(x)$

Focus on how $g(x)$ is built from $f$ . Since $g(x)=5-3f(2x+4)$ , what happens to $g(x)$ when $f(2x+4)$ gets larger or smaller?

Rename the inside of $f$

Try setting $t = 2x+4$ so that you can think about minimizing $f(t)$ instead of $f(2x+4)$ . Then write $f(t)=2^{t-1}+2^{1-t}$ and see if you can simplify the second term.

Use a substitution for the powers of 2

After you notice that $2^{1-t}=1/2^{t-1}$ , let $u = 2^{t-1}$ and rewrite $f(t)$ in terms of $u$ . Then ask: for $u>0$ , when is $u + 1/u$ as small as possible?

Convert back to $x$

Once you know the value of $t$ that makes $f(t)$ smallest, use the equation $t = 2x+4$ to solve for the corresponding value of $x$ that gives the maximum of $g(x)$ .

Desmos Guide

Enter the function $f$

In Desmos, type f(x) = 2^(x-1) + 2^(1-x) to define the function $f$ .

Enter the function $g$

On a new line, type g(x) = 5 - 3*f(2*x + 4) so that Desmos graphs the composite function $g(x)$ .

Find the maximum of $g(x)$

Adjust the viewing window (for example, set $x$ from about -4 to 2) until you can clearly see the highest point of the graph of $g(x)$ . Click on that highest point; Desmos will display its coordinates. The $x$ -coordinate of this maximum point is the value of $x$ you should choose from the answer options.

Step-by-step Explanation

Connect maximizing $g(x)$ to minimizing $f(2x+4)$

The function $g$ is defined by

g(x)=5-3f(2x+4).

Because the coefficient of $f(2x+4)$ is $-3$ (negative), as $f(2x+4)$ gets larger, $g(x)$ gets smaller, and as $f(2x+4)$ gets smaller, $g(x)$ gets larger.

So, to maximize $g(x)$ , we need to minimize the value of $f(2x+4)$ .

Introduce a simpler variable for the input of $f$

Let

t = 2x+4.

Then we are trying to minimize $f(t)$ , where

f(t)=2^{t-1}+2^{1-t}.

Once we find the value of $t$ that minimizes $f(t)$ , we will convert that $t$ back to $x$ using $t=2x+4$ .

Rewrite $f(t)$ in a more convenient form

Notice that

2^{1-t} = 2^{-(t-1)} = \frac{1}{2^{t-1}}.

So if we let

u = 2^{t-1},

then $u>0$ and

f(t) = 2^{t-1} + 2^{1-t} = u + \frac{1}{u}.

Now the problem becomes: for $u>0$ , when is $u + \dfrac{1}{u}$ as small as possible?

Find the minimum of $u + 1/u$ and solve for $x$

For any positive number $u$ , the expression $u + \dfrac{1}{u}$ is minimized at $u=1$ (you can see this from symmetry, by AM-GM, or by checking values like $u=\tfrac12,1,2$ and noticing it increases when you move away from 1). At $u=1$ , we get

u + \frac{1}{u} = 1 + 1 = 2,

and this is the smallest possible value.

Since $u = 2^{t-1}$ , the condition $u=1$ means

2^{t-1}=1 \quad\Rightarrow\quad t-1=0 \quad\Rightarrow\quad t=1.

Recall that $t=2x+4$ , so

2x+4 = 1 \quad\Rightarrow\quad 2x = -3 \quad\Rightarrow\quad x = -\frac{3}{2}.

Thus $g(x)$ reaches its maximum at $x = -\frac{3}{2}$ .

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation