Nested Function Value Count
If \[f(x) =
\begin{cases}
2x-5 &\quad \text{if } x \ge 3, \\
-x + 5 &\quad \text{if } x < 3,
\end{cases}
\]then for how many values of $ x $ is $ f(f(x)) = 3 $?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ y = f(x) $. Then, $ f(f(x)) = f(y) = 3 $, so either $ 2y - 5 = 3 $ or $ -y + 5 = 3 $. If $ 2y - 5 = 3 $, then $ y = 4 $. Note that $ 4 \ge 3 $, so $ f(4) = 3 $. If $ -y + 5 = 3 $, then $ y = 2 $. Note that $ 2 < 3 $, so $ f(2) = 3 $. So the two possible values of $ y $ are 2 and 4.
Now, we solve the equation $ f(x) = 2 $. In this case, either $ 2x - 5 = 2 $ or $ -x + 5 = 2 $. If $ 2x - 5 = 2 $, then $ x = 7/2 $. Note that $ 7/2 \ge 3 $, so $ f(7/2) = 2 $. If $ -x + 5 = 2 $, then $ x = 3 $. But $ 3 \ge 3 $, so $ f(3) = 2 \cdot 3 - 5 = 1 $, which is not 2.
Next, we solve the equation $ f(x) = 4 $. In this case, either $ 2x - 5 = 4 $ or $ -x + 5 = 4 $. If $ 2x - 5 = 4 $, then $ x = 9/2 $. Note that $ 9/2 \ge 3 $, so $ f(9/2) = 3 $. If $ -x + 5 = 4 $, then $ x = 1 $. Note that $ 1 < 3 $, so $ f(1) = 4 $.
Therefore, there are $ \boxed{3} $ solutions to $ f(f(x)) = 3 $, namely $ x = 1 $, 7/2, and 9/2.
Freeflows