The function is defined by For what value of does attain its minimum?

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SAT Nonlinear Functions Question 129 | Free SAT Prep | Aniko

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Question 129·Hard·Nonlinear Functions

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The function $g$ is defined by

g(x)=2^{x-5}+2^{5-x}

For what value of $x$ does $g(x)$ attain its minimum?

Your answer

(Express the answer as an integer)

For exponential expressions of the form $2^{x-c}+2^{c-x}$ on the SAT, look for symmetry around the constant $c$ . A quick way is to substitute $t=x-c$ so the function becomes $2^t+2^{-t}$ , then let $a=2^t$ to get $a+1/a$ with $a>0$ . Use the fact (from AM-GM or basic reasoning) that $a+1/a$ is always at least 2 and achieves this smallest value when $a=1$ . Finally, translate that condition back to $x$ . Recognizing this pattern saves time and avoids trial-and-error with multiple $x$ -values.

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Notice the symmetry in the exponents

Look at the exponents $x-5$ and $5-x$ . How are they related to each other? Try expressing both in terms of $x-5$ (or $5-x$ ) so they look like opposites.

Use a substitution to simplify the expression

Let $t=x-5$ . Rewrite $g(x)$ in terms of $t$ . This should give you a function involving $2^t$ and $2^{-t}$ .

Turn it into a simpler algebraic expression

After substituting, let $a=2^t$ . Then your expression becomes $a+\dfrac{1}{a}$ with $a>0$ . Think about how small $a+\dfrac{1}{a}$ can be for positive $a$ .

Find when the lower bound is reached

Once you know a lower bound for $a+\dfrac{1}{a}$ , ask: for what value of $a$ is that bound actually achieved? Then translate that back into $t$ and then into $x$ .

Desmos Guide

Enter the function

In Desmos, type g(x) = 2^(x-5) + 2^(5-x) into an expression line to graph the function.

View the important part of the graph

Use zoom in/out so you can clearly see the bottom of the curve near the center of the graph; you should see a smooth U-shaped graph (like a shallow valley).

Find the minimum point

Click on the graph near the lowest point of the curve; Desmos will show a point labeled with coordinates $(x,y)$ at the minimum. The $x$ -coordinate of this point is the value of $x$ where $g(x)$ attains its minimum.

Step-by-step Explanation

Center the expression around 5

The exponents in $g(x)=2^{x-5}+2^{5-x}$ are $x-5$ and $5-x$ , which are opposites of each other. Let $t=x-5$ . Then $5-x=-(x-5)=-t$ . So we can rewrite the function as

g(x)=2^{t}+2^{-t}.

Now we are looking for the value of $t$ (and therefore $x$ ) that makes $2^{t}+2^{-t}$ as small as possible.

Rename the expression in terms of a single positive variable

Let $a=2^t$ . Since a power of 2 is always positive, we know $a>0$ . Then $2^{-t} = \dfrac{1}{2^t} = \dfrac{1}{a}$ . So

2^{t}+2^{-t} = a + \frac{1}{a}, \quad \text{where } a>0.

Now the problem becomes: for $a>0$ , how small can $a+\dfrac{1}{a}$ be?

Find a lower bound for $a+1/a$

Use the AM-GM inequality for positive numbers $a$ and $\dfrac{1}{a}$ :

The arithmetic mean is

\frac{a+\frac{1}{a}}{2}.

The geometric mean is

\sqrt{a\cdot \frac{1}{a}} = 1.

AM-GM says arithmetic mean $\ge$ geometric mean, so

\frac{a+\frac{1}{a}}{2} \ge 1.

Multiply both sides by 2:

a+\frac{1}{a} \ge 2.

So $g(x)=a+\dfrac{1}{a}$ can never be less than 2.

Determine when the minimum occurs and convert back to x

The inequality $a+\dfrac{1}{a} \ge 2$ becomes an equality only when $a$ and $\dfrac{1}{a}$ are equal, so $a=1$ . We defined $a=2^t$ , so we need $2^t=1$ , which happens when $t=0$ . Recall $t=x-5$ , so $x-5=0$ and therefore $x=5$ . Thus, $g(x)$ attains its minimum when $x=5$ .

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Desmos Guide

Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation