### Instruction: Let
\[f(x) = \left\{
\begin{array}{cl}
x + 3 & \text{if $x < 20$}, \\
2x - 2 & \text{if $x \ge 20$}.
\end{array}
\right.\]Find $f^{-1}(7) + f^{-1}(46).$.### Response: I notice that the function $f$ is a piecewise function, defined by two different linear equations for different intervals of $x$.
To find $f^{-1}(7)$, I need to figure out which equation satisfies $f(x) = 7$.
I try the first equation, $x + 3 = 7$, and I get $x = 4$.
I check if this is consistent with the condition $x < 20$, and it is, so $f^{-1}(7) = 4$.
To find $f^{-1}(46)$, I need to figure out which equation satisfies $f(x) = 46$.
I try the second equation, $2x - 2 = 46$, and I get $x = 24$.
I check if this is consistent. With the condition $x \ge 20$, and it is, so $f^{-1}(46) = 24$.
To find the sum, I just add the two values of $x$ that I found.
I get $f^{-1}(7) + f^{-1}(46) = 4 + 24 = 28$.