The function is defined for all real numbers by For what value of does attain its minimum value?

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SAT Nonlinear Functions Question 111 | Free SAT Prep | Aniko

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Question 111·Hard·Nonlinear Functions

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The function $h$ is defined for all real numbers $x$ by

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

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

Your answer

(Express the answer as an integer)

When a function involves terms like $a^{x+k}$ and $a^{m-x}$ , first use exponent rules to rewrite them in terms of $a^x$ and $a^{-x}$ and factor out any positive constant (which doesn’t affect where the minimum occurs). Then use a substitution such as $t=a^x$ (so $t>0$ ) to turn the problem into minimizing a simpler expression like $t+1/t$ . Apply tools like AM-GM or basic calculus to find the minimum of that simpler expression, and finally translate back to $x$ . This approach is fast and avoids messy trial-and-error with the original exponents.

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Look for a simpler form of the expression

Use exponent rules to rewrite $2^{x+1}$ and $2^{1-x}$ so that both terms involve $2^x$ or $2^{-x}$ ; factor out any common positive constant.

Use a substitution

After simplifying, try letting $t=2^x$ so that the expression becomes something like $t+\dfrac{1}{t}$ with the condition $t>0$ .

Minimize the new expression

For positive $t$ , think about how to find the minimum of $t+\dfrac{1}{t}$ . You can use inequalities like AM-GM, or you can reason about its graph or derivative.

Desmos Guide

Enter the function

In Desmos, type h(x) = 2^(x+1) + 2^(1-x) to graph the function.

Locate the minimum point

Zoom or adjust the viewing window so you can clearly see the lowest point of the graph. Then tap/click on that lowest point (or use the Desmos minimum function) and note the x-coordinate where the graph reaches its smallest y-value; that x-value is the solution.

Step-by-step Explanation

Rewrite the function using exponent rules

Start by simplifying each term using $2^{a+b}=2^a\cdot 2^b$ and $2^{-x}=\frac{1}{2^x}$ .

$2^{x+1}=2^x\cdot 2^1=2\cdot 2^x$
$2^{1-x}=2^1\cdot 2^{-x}=2\cdot 2^{-x}$

h(x)=2^{x+1}+2^{1-x}=2\cdot 2^x+2\cdot 2^{-x}=2\bigl(2^x+2^{-x}\bigr).

Because the factor $2$ is positive, minimizing $h(x)$ is the same as minimizing $2^x+2^{-x}$ .

Use a substitution to simplify the expression

Let $t=2^x$ . Since $2^x$ is always positive, we know $t>0$ .

Then $2^{-x}=\dfrac{1}{2^x}=\dfrac{1}{t}$ .

So the part we need to minimize becomes

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

Now the problem is: for positive $t$ , when is $t+\dfrac{1}{t}$ as small as possible?

Find the minimum of $t+\dfrac{1}{t}$ for $t>0$

Use the AM-GM inequality, which says that for positive numbers $a$ and $b$ ,

\frac{a+b}{2} \ge \sqrt{ab},

with equality only when $a=b$ .

Here, take $a=t$ and $b=\dfrac{1}{t}$ (both positive):

\frac{t+\dfrac{1}{t}}{2} \ge \sqrt{t\cdot \dfrac{1}{t}} = \sqrt{1}=1.

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

Equality (and thus the minimum value) happens when $t=\dfrac{1}{t}$ , which means $t^2=1$ and, since $t>0$ , we must have $t=1$ .

Translate back to $x$ and answer the question

We found that the expression $t+\dfrac{1}{t}$ is minimized when $t=1$ .

From our substitution, $t=2^x$ , so

2^x=1.

The exponential function $2^x$ equals $1$ only when $x=0$ , because $2^0=1$ .

Therefore, $h(x)$ attains its minimum value when $x=\boxed{0}$ .

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Step-by-step Explanation

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