Answer
(a)
$f^{-1}_1=40-4\sqrt t$ - When water is drained.
$f^{-1}_2=40+4\sqrt t$ - When water is pumped in (by the rate of drain).
(b)
$f^{-1}(15)\approx 24.5$ (When water drains)
It means, that when there are $15$ gallons of water left in the tank $24.5$ minutes have passed.
Work Step by Step
$f(t)=100(1-\frac{t}{40})^2$
(a) To find $f^{-1}$ we will follow some steps, which will be just for purpose of better visualization.
At first, write the expression in terms of $y$ and $t$ (originally $x$):
$y=100(1-\frac{t}{40})^2$
Then replace $y$ by $t$ and vice versa:
$t=100(1-\frac{y}{40})^2$
And at last, solve for $y$:
$t=100(1-\frac{y}{20}+\frac{y^2}{1600})$
$t=100-5y+\frac{y^2}{16}$
$y^2-80y+1600-16t=0$
$D=6400-6400+64t=64t$
$y_1=\frac{80-8\sqrt{t}}{2}=40-4\sqrt t$
$y_2=\frac{80+8\sqrt t}{2}=40+4\sqrt t$
So, we have got that $f^{-1}$ has two solution:
$f^{-1}_1=40-4\sqrt t$
$f^{-1}_2=40+4\sqrt t$
$f^{-1}_1$ represents $y$ (time passed) for given $t$ amount of water (volume) left in the tank.
$f^{-1}_2$ represents almost the same, but instead of draining water, we have it being filled up, by the same rate as it drains.
(b) (In this case we will use $f^{-1}_1$, as the situation mentioned is when water drains)
$f^{-1}(15)=40-4\sqrt{15}\approx 24.5$
It means, that when there are $15$ gallons of water left in the tank $24.5$ minutes are passed.
