Question 598·Hard·Math

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

I. $f(x)=14\cdot7^{-(x+1)}$

II. $g(x)=2\cdot7^{-x}$

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

In exponential decay, increasing the input makes the output smaller. With $x\ge0$ , the largest output is at $x=0$ . Evaluate each function there in Desmos, then check whether that value is actually written as a constant or coefficient. A number in front of a power isn't always the value at $x=0$ when the exponent includes a shift.

Hints

Hint 1
Each increase of $1$ in $x$ multiplies both outputs by $\frac17$ , a factor below $1$ . This is exponential decay. Which allowed input gives the largest output?
Hint 2
At $x=0$ , the exponent in I is $-(x+1)$ , not $-x$ . Evaluate both functions at $0$ rather than assuming the numbers in front are their maximum values.
Hint 3
A coefficient is a number multiplying an expression. Once you know each maximum, ask whether that number is written in the equation as a constant or coefficient. Having a maximum and displaying it are different things.

Step-by-step

Approach 1: Evaluate at the allowed endpoint
Step 1
Locate both maximum values
Each increase of $1$ in $x$ lowers both exponents by $1$ , so each output is multiplied by $7^{-1}=\frac17$ . That factor is below $1$ , so both functions decrease as $x$ grows. Since $x=0$ is allowed, both reach their maximum there.
Step 2
Evaluate I at zero
Type $f(x)=14\cdot7^{-(x+1)}$ and then $f(0)$ in Desmos. It shows $2$ , so I's maximum is $2$ , not the $14$ in front. The $+1$ inside the exponent means substituting $0$ does not make that exponent $0$ .
Step 3
Evaluate II and check which equation displays its maximum
Keep those lines and add $g(x)=2\cdot7^{-x}$ and $g(0)$ . Desmos shows $2$ , so II's maximum is $2$ . Here the exponent is $0$ at $x=0$ , making the written coefficient $2$ the starting output. Both maximum values are $2$ . I shows $14$ as its coefficient, while II shows $2$ as its coefficient. Find the maximum first, then check whether that value is written in the equation. So only II displays its maximum as a coefficient. Choice B.

Approach 2: Show why the functions match
Step 1
Remove the parentheses in I's exponent
Distribute the negative sign across $x+1$ :
$-(x+1)=-x-1.$
Step 2
Separate the extra power of seven
Multiplying powers with the same base adds their exponents, so split I's power:
$f(x)=14\cdot7^{-x}\cdot7^{-1}.$
Step 3
Rewrite the negative power
A negative exponent means a reciprocal, so replace $7^{-1}$ with $\frac17$ :
$f(x)=14\cdot7^{-x}\cdot\frac17.$
Step 4
Compare the identical functions
Divide $14$ by $7$ :
$f(x)=2\cdot7^{-x}=g(x).$
The functions have the same maximum, $2$ at $x=0$ , but only II writes $2$ as its coefficient. So II only displays its maximum. Choice B.

Question 598·Hard·Math

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