Question 865·Hard·Math

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For $x \ge 0$ , the functions are defined as follows.

I. $f(x)=4^{-(x+1)}$

II. $g(x)=\dfrac14\,4^{-x}$

Which of these equations displays, as a constant or coefficient, the maximum value of the function it defines, where $x \ge 0$ ?

In an exponential decay function, outputs fall as the input rises. Use the allowed inputs to locate the maximum, then evaluate each function there in Desmos. Finally, inspect each equation as written: rewriting a function can reveal a maximum value without making that value visible in the equation you were given.

Hints

Hint 1
The domain is the set of allowed inputs. Here it includes $0$ . As $x$ rises, what happens to a power of $4$ when its exponent becomes more negative?
Hint 2
A coefficient is a number multiplying an expression. After finding the greatest output, look for that number outside the power in each equation as written.

Step-by-step

Find the maximum, then inspect the written forms
Step 1
Locate each maximum
As $x$ grows, both $-x$ and $-(x+1)$ decrease. Powers of $4$ get smaller when their exponents decrease, and multiplying by the positive $\frac14$ doesn't change that. So both functions decrease for $x\ge0$ . Their maximum, or greatest output, occurs at the allowed input $x=0$ .
Step 2
Evaluate both functions at zero
Type both function definitions in Desmos, then type $f(0)$ and $g(0)$ . Desmos shows $0.25$ for each, which is $\frac14$ . So the two functions have the same maximum value.
Step 3
Check what each equation displays
The question asks which equation displays $\frac14$ as a constant or coefficient, a number multiplying an expression. In II, $\frac14$ visibly multiplies $4^{-x}$ . You could rewrite I as $\frac14\,4^{-x}$ , but I isn't written that way; its maximum is hidden in the exponent. So only II displays the maximum value. Choice B.

Question 865·Hard·Math

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