Let \[f(x) = \left\{
\begin{array}{cl} x^2-4 & \text{ if }x < 7, \\
x-13 & \text{ if } x \geq 7.
\end{array}
\right.\] What is $f(f(f(17)))$?
We approach this problem by working from the inside out, so we first find $f(17)$. Since $17 \geq 7$, $f(17) = 17 - 13 = 4$. Next, $4<7$, so $f(f(17))=f(4)=(4)^2-4=12$. Finally we have that since $12 \geq 7$, $f(f(f(17)))=f(12)=12-13=\boxed{-1}$.
-1
Level 2
Algebra
Problem: Let \[f(x) = \left\{
\begin{array}{cl} x^2-4 & \text{ if }x < 7, \\
x-13 & \text{ if } x \geq 7.
\end{array}
\right.\] What is $f(f(f(17)))$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( f(f(f(17))) \), we first need to evaluate \( f(17) \). Since \( 17 \geq 7 \), we use the second case of the piecewise function: 

\[
f(17) = 17 - 13 = 4.
\]

Next, we evaluate \( f(f(17)) = f(4) \). Since \( 4 < 7 \), we use the first case of the piecewise function:

\[
f(4) = 4^2 - 4 = 16 - 4 = 12.
\]

Now, we evaluate \( f(f(f(17))) = f(12) \). Since \( 12 \geq 7 \), we again use the second case of the piecewise function:

\[
f(12) = 12 - 13 = -1.
\]

Thus, the final answer is \( -1 \).<|END_OF_LATENT|> The final answer is $-1$.