Let
\[f(x) = \left\{
\begin{array}{cl}
x + 3 & \text{if $x < 20$}, \\
2x - 2 & \text{if $x \ge 20$}.
\end{array}
\right.\]Find $f^{-1}(7) + f^{-1}(46).$
To find $f^{-1}(7),$ we try solving $f(x) = 7$ on each piece.

If $x + 3 = 7$, then $x = 4,$ which satisfies $x < 20.$  If $2x - 2 = 7,$ then $x = \frac{9}{2},$ which does not satisfy $x \ge 20,$ so $f^{-1}(7) = 4.$

Similarly, $x + 3 = 46,$ then $x = 43,$ which does not satisfy $x < 20.$  If $2x - 2= 46,$ then $x = 24,$ which satisfies $x \ge 20,$ so $f^{-1}(46) = 24.$

Hence, $f^{-1}(7) + f^{-1}(46) = 4 + 24 = \boxed{28}.$
28
Level 3
Intermediate Algebra
Problem: Let
\[f(x) = \left\{
\begin{array}{cl}
x + 3 & \text{if $x < 20$}, \\
2x - 2 & \text{if $x \ge 20$}.
\end{array}
\right.\]Find $f^{-1}(7) + f^{-1}(46).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( f^{-1}(7) \), we need to determine which piece of the piecewise function applies. Since \( 7 < 20 \), we use the first case: \( f(x) = x + 3 \). Setting this equal to 7 gives:

\[
x + 3 = 7
\]

Solving for \( x \):

\[
x = 7 - 3 = 4
\]

Thus, \( f^{-1}(7) = 4 \).

Next, we find \( f^{-1}(46) \). Since \( 46 \ge 20 \), we use the second case: \( f(x) = 2x - 2 \). Setting this equal to 46 gives:

\[
2x - 2 = 46
\]

Solving for \( x \):

\[
2x = 46 + 2 = 48
\]
\[
x = \frac{48}{2} = 24
\]

Thus, \( f^{-1}(46) = 24 \).

Now, we add the two results together:

\[
f^{-1}(7) + f^{-1}(46) = 4 + 24 = 28
\]

The final answer is \( 28 \).<|END_OF_LATENT|> The final answer is $28$.