Answer
(a) $f(10)=100+0.2 \cdot 10=102$ thousand dollars.
b) $f^{-1}(200)=500$.
c) $f^{-1}(C)=5 C-500$
Work Step by Step
(a) $f(10)=100+0.2 \cdot 10=102$ thousand dollars, the cost of producing 10 kg of the chemical.
(b) $f^{-1}(200)$ is the quantity of the chemical which can be produced for 200 thousand dollars. Since
$$
\begin{aligned}
200 & =100+0.2 q \\
0.2 q & =100 \\
q & =\frac{100}{0.2}=500 \mathrm{~kg}
\end{aligned}
$$
we have $f^{-1}(200)=500$.
(c) To find $f^{-1}(C)$, solve for $q$ :
$$
\begin{aligned}
C & =100+0.2 q \\
0.2 q & =C-100 \\
q & =\frac{C}{0.2}-\frac{100}{0.2}=5 C-500 \\
f^{-1}(C) & =5 C-500
\end{aligned}
$$
