Question 47·200 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 integer constants and $a>b>0$ . If the graph of each function is considered only for $x\ge 0$ , which of the following equations displays, as a constant or a coefficient, the maximum value of the graph of the corresponding function?

I. $f(x)=a-(x-b)^2$

II. $g(x)=a-b(0.9)^x$

A maximum value must occur at an allowed input, not merely be a height the graph approaches. A negative squared term can show a parabola’s peak, while an exponential term can shrink toward zero without ever becoming zero. Check the input restriction before accepting a peak. Approaching a height is not the same as reaching it.

Hints

Hint 1
A squared term such as $(x-b)^2$ cannot be negative. Subtracting it from $a$ gives the largest possible output when the square is $0$ . What input makes that happen?
Hint 2
The domain is the set of allowed inputs. Since $b>0$ , $x=b$ is allowed under $x\ge 0$ . Does the quadratic reach its largest possible output there?
Hint 3
For every finite $x$ , $(0.9)^x$ stays positive, even as it shrinks toward $0$ . In $g(x)$ , is the amount subtracted from $a$ ever zero?

Step-by-step

Check whether each function reaches a highest value
Step 1
Bound the quadratic’s output
No Desmos needed. The square and the shrinking exponential show what happens at every allowed input. A square is never negative, so $(x-b)^2\ge 0$ . Subtracting it means $f(x)$ cannot exceed $a$ :
$f(x)=a-(x-b)^2\le a.$
Step 2
Check that the quadratic reaches the bound
At $x=b$ , the square is $0$ , so $f(b)=a$ . The vertex, the parabola’s turning point, is at $(b,a)$ . Because $b>0$ , $x=b$ is allowed by $x\ge 0$ . So I displays its maximum value as the constant $a$ .
Step 3
Check whether the exponential reaches $a$
In II, $(0.9)^x$ is positive for every finite $x$ . Since $b>0$ , the amount $b(0.9)^x$ is positive too. So $a$ is never an output:
$g(x)=a-b(0.9)^x<a.$
Step 4
Decide whether the exponential has another maximum
Because $0<0.9<1$ , increasing $x$ makes $(0.9)^x$ smaller. That subtracts less from $a$ , so $g(x+1)>g(x)$ . For every allowed input, $x+1$ is also allowed and gives a higher output. II has no maximum: its graph rises toward $a$ but never reaches it. Only I displays a maximum value. Choice A.

Question 47·200 Super-Hard SAT Math Questions·Advanced Math

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