Super-Hard SAT Math Question 45 | Free SAT Prep | Aniko

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Question 45·200 Super-Hard SAT Math Questions·Advanced Math

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The functions $f$ and $g$ are defined by the given equations, where $x \ge 0$ . Which of the following equations displays, as a constant or coefficient, the maximum value of the function it defines, where $x \ge 0$ ?

I. $f(x)=64\left(\frac{4}{5}\right)^{x-2}$

II. $g(x)=64\left(\frac{5}{4}\right)^2\left(\frac{4}{5}\right)^x$

When you see an exponential function with a base between 0 and 1, immediately recognize it decreases as $x$ increases. On a restricted domain like $x\ge 0$ , the maximum is at the smallest allowed $x$ (usually $x=0$ ). Then check whether the function is written in the standard form $A\cdot b^x$ ; if so, the maximum equals the coefficient $A$ . If the exponent is shifted (like $x-2$ or $x+3$ ), rewrite using exponent rules to see what the true coefficient at $b^x$ would be.

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Look at the base

Since the base is between 0 and 1, think about whether the function increases or decreases as $x$ increases.

Where is the maximum on $x\ge 0$ ?

If the function is decreasing, the maximum happens at the smallest allowed $x$ value.

What form makes the maximum a coefficient?

Compare each equation to the form $A\left(\frac{4}{5}\right)^x$ . In that form, what does the coefficient $A$ represent when $x=0$ ?

Desmos Guide

Graph both functions

Enter

$y_1=64\left(\frac{4}{5}\right)^{x-2}$
$y_2=64\left(\frac{5}{4}\right)^2\left(\frac{4}{5}\right)^x$

Make sure you can see the graphs for $x\ge 0$ .

Confirm where the maximum occurs

Use the table for each function and evaluate at $x=0,1,2$ . You should see the values decrease as $x$ increases, so the maximum occurs at $x=0$ .

Compare the maximum value to what is shown in each equation

Note the value at $x=0$ (the maximum). Then check which equation is written as a coefficient times $\left(\frac{4}{5}\right)^x$ so that the coefficient equals that $x=0$ value.

Make the selection

Choose the option for which the equation’s displayed coefficient matches the maximum value at $x=0$ .

Step-by-step Explanation

Use the base to locate the maximum

In both equations, the exponential base is $\frac{4}{5}$ , and $0<\frac{4}{5}<1$ .

So as the exponent increases, the value of $\left(\frac{4}{5}\right)^{\text{exponent}}$ decreases. Therefore, on the domain $x\ge 0$ , the maximum value occurs at the smallest allowed $x$ , which is $x=0$ .

Find each function’s maximum value (its value at $x=0$ )

Evaluate each function at $x=0$ .

For I:

\begin{aligned} f(0) &= 64\left(\frac{4}{5}\right)^{-2} \\ &= 64\left(\frac{5}{4}\right)^2 \end{aligned}

For II:

\begin{aligned} g(0) &= 64\left(\frac{5}{4}\right)^2\left(\frac{4}{5}\right)^0 \\ &= 64\left(\frac{5}{4}\right)^2 \end{aligned}

So both functions have the same maximum value, $64\left(\frac{5}{4}\right)^2$ (which equals $100$ ).

Check whether that maximum is displayed as a constant or coefficient

An equation “displays” the maximum as a constant or coefficient if it is written like

A\left(\frac{4}{5}\right)^x

because then the coefficient $A$ equals the value at $x=0$ (and thus the maximum for $x\ge 0$ ).

In II, the function is already written as

\left[64\left(\frac{5}{4}\right)^2\right]\left(\frac{4}{5}\right)^x

so the maximum value is the coefficient $64\left(\frac{5}{4}\right)^2$ .

In I, the function is written as $64\left(\frac{4}{5}\right)^{x-2}$ , so the coefficient shown is $64$ , which is not the maximum value.

Select the correct choice

Only equation II displays, as a constant/coefficient, the maximum value of the function it defines.

Correct answer: II only

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Step-by-step Explanation

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Step-by-step Explanation