Question 59·200 Super-Hard SAT Math Questions·Advanced Math

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The functions $f$ and $g$ are defined by the equations shown, where $a$ and $b$ are positive integers and $a>b$ . If $y=f(x)$ and $y=g(x)$ are graphed in the $xy$ -plane, which of the following equations displays, as a constant or a coefficient, the maximum value of the graph of the corresponding function for $x\ge 0$ ?

I. $f(x)=a(1.15)^{-x}$

II. $g(x)=a-b(1.15)^{-x}$

A maximum is the greatest output a graph actually takes. With a base above $1$ , a negative exponent makes a positive factor shrink as $x$ moves right. Sketch an allowed example in Desmos, then use that factor to check the included endpoint and whether a rising graph ever reaches the height it approaches. An asymptote is not automatically a maximum.

Hints

Hint 1
The domain is the set of allowed inputs. To see the shapes, graph an example such as $a=3$ and $b=1$ using only allowed inputs. What happens to each curve as you move right?
Hint 2
A negative exponent means reciprocal: $(1.15)^{-x}=1/(1.15)^x$ . For $x\ge0$ , this factor starts at $1$ , then stays positive while shrinking. What does that tell you about the highest output each graph can take?
Hint 3
A horizontal asymptote is a height a graph approaches. If a positive amount is always subtracted from $a$ , can the resulting output ever equal $a$ ?

Step-by-step

Graph the shapes, then check which height is reached
Step 1
See the directions of both graphs
Type $a=3$ , $b=1$ , and both functions in Desmos. Add $\{x\ge0\}$ to each function to show its allowed inputs. These values satisfy $a>b>0$ . Desmos shows $f$ falling and $g$ rising as you move right. That's a shape check, not proof for every allowed $a$ and $b$ .
Step 2
Bound the factor both functions use
Rewrite the negative exponent as a reciprocal:
$(1.15)^{-x}=\frac{1}{(1.15)^x}.$
At $x=0$ , the denominator is $1$ . For larger $x$ , it grows because $1.15>1$ . So for every allowed input:
$0<(1.15)^{-x}\le1.$
The factor shrinks toward $0$ , but is never zero.
Step 3
Find an upper bound for f
An upper bound is a value the output cannot exceed. Multiply the factor's bound by $a>0$ :
$f(x)=a(1.15)^{-x}\le a.$
So no output of $f$ exceeds $a$ . Now check whether the graph takes that value.
Step 4
Show that f reaches its upper bound
Add $f(0)$ in Desmos. It shows $3$ for the example, matching its value of $a$ . In general, $x=0$ is included, and the zero-exponent rule gives:
$f(0)=a(1)=a.$
Since $f$ never exceeds $a$ and takes that value at $0$ , its maximum is $a$ .
Step 5
Check whether g reaches a maximum
For $g$ , $b>0$ and the exponential factor is strictly positive. Subtracting it from $a$ gives:
$g(x)=a-b(1.15)^{-x}<a.$
As $x$ grows, that factor keeps shrinking, so $g(x)$ keeps rising toward $a$ without reaching it. A height approached but never reached is not a maximum. Only I displays its graph's maximum as a coefficient. Choice A.

Question 59·200 Super-Hard SAT Math Questions·Advanced Math

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