+206 If
\[f(x) =
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
\]
then for how many values of $x$ is $f(f(x)) = 5$?
+118687 \(f(x) = \begin{cases} x^2-4 &\quad \text{if } x \ge -4, \\ x + 3 &\quad \text{otherwise}, \end{cases}\)
then for how many values of \(x\) is \(f(f(x)) = 5\) ?
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You need to look at 2 different cases.
first assume f(x) is less than -4
solve
5=x+3
x=-2
-2 is not less than -4 so no solutions here.
now assume f(x) >= -4
\(5=x^2-4\\ 9=x^2\\ x=\pm3\)
Both of these are bigger than -4 so they both could have answers. That is, for this problem
f(x) could be +3 or -3
Now you need to look at each of these for each of the scenarios x<-4 and x>=-4 and see how many answers you get that are valid.
