Inverse functions can be solved in the following way (example)

If we have a function $f(x) = y = mx + c$
Then $f^{-1}(x)$ can be solved by interchanging x and y values i.e. $f^{-1}(x) = x = my + c$

NOTE: Do not confuse inverse of function with the reciprocal of function
$\frac{1}{f(x)}$ is the reciprocal not the inverse!

Also, Inverse is written as $f^{-1}(x)$ whereas, Reciprocal would be $[f(x)]^{-1}$

Now we are given,
$f(x) = 3x - 15$
So, $f^{-1}(-3) = 3y - 15$
$-3 = 3y - 15$
$y = 4$

$g(x) = (x+3)^2 + 3$
$g^{-1}(4) = (y+3)^2 + 3$
$4 = (y+3)^2 + 3$
$1 = (y+3)^2$
$y = -2$ or $-4$

Col. A: $4$
Col. B: $-2$ or $-4$

Hence, option A

Let's learn to find Inverse function $f^{-1}(x)$ first

To find $f^{-1}(x)$ we need to write x in terms of f(x)
$f(x)=3x-15$
=> 3x = f(x) + 15
=> x = $\frac{f(x)}{3} + 5$
Now replace x with $f^{-1}(x)$ we get
$f^{-1}(x)$ = $\frac{f(f^{-1}(x))}{3}$ + 5
=> $f^{-1}(x)$ = $\frac{x}{3}$ + 5

Let's find $g^{-1}(x)$ similarly
$g(x)=(x+3)^2+3$
$(x+3)^2 = g(x) - 3$
Now there will be two cases
Case 1:
x+3 = +$\sqrt{g(x) - 3}$
=> x = +$\sqrt{g(x) - 3}$ - 3
Replace x with $g^{-1}(x)$ we get
$g^{-1}(x)$ = $\sqrt{g(g^{-1}(x)) - 3}$ - 3
=> $g^{-1}(x)$ = $\sqrt{x - 3}$ - 3

Case 2:
x+3 = -$\sqrt{g(x) - 3}$
=> x = -$\sqrt{g(x) - 3}$ - 3
Replace x with $g^{-1}(x)$ we get
$g^{-1}(x)$ = -$\sqrt{g(g^{-1}(x)) - 3}$ - 3
=> $g^{-1}(x)$ = -$\sqrt{x - 3}$ - 3


Quantity A: $f^{-1}(-3)$
=> $f^{-1}(x)$ = $\frac{x}{3}$ + 5
=> $f^{-1}(-3)$ = $\frac{-3}{3}$ + 5 = -1 + 5 = 4

Quantity B: $g^{-1}(4)$
Case 1
=> $g^{-1}(x)$ = $\sqrt{x - 3}$ - 3
=> $g^{-1}(4)$ = $\sqrt{4 - 3}$ - 3 = 1-3 = -2

Case 2
=> $g^{-1}(x)$ = -$\sqrt{x - 3}$ - 3
=> $g^{-1}(4)$ = -$\sqrt{4 - 3}$ - 3 = -1-3 = -4


Clearly, Quantity A(4) > Quantity B(-2 or -4)

So, Answer will be A
Hope it helps!

To learn more about Functions watch the following video

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.