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I saw this problem a few days ago and still haven't cracked it. Therefore, I thought I might ask you all for help.
The graph of $f(x)$ has four roots on the interval $[-5,5]$. How many different roots does $f(f(x))$ have? Here is a picture of the graph of $f(x)$:
We stumbled across this problem a week ago during math club and we were unable to solve it; our teacher didn't know the answer, either. Any help would thus be greatly appreciated!
The roots of $f(x)$ are $$x=-4,-2,2,4$$, according to your graph, so for $f(f(x)) = 0$, we have $$f(x) = -4,-2,2,4$$ because $f(x)$ took the place of $x$. So, we need to examine when $f(x) = -4,-2,2,4$.
From the graph we see that $f(x)$ is never $-4$
From the graph we see that $f(x)=-2$ at $2$ values of $x$.
From the graph we see that $f(x) = 2$ at $4$ values of $x$
From the graph we see that $f(x) = 4$ at $2$ values of $x$
Thus, that makes $8$ values of $x$ total that $f(f(x)) = 0 $, so it has $8$ roots.
Notice that $$ f(f(x)) = 0 $$ if and only if $$ f(x) \in S=\{-4,-2,2,4\} , $$ since $S$ is the set of the roots of $f$.
So you should find how many times $f(x) = y$, with $y$ being a value in $S$. I suggest to draw four horizontal lines, respectively on $y=-4$ (which lies outside the graphics), $y=-2$, $y=2 and $y=4$.
You can notice that the line $y=-4$ has zero intersections with the graph of the function $f$; the line $y=-2$ has 2 intersections; the line $y=2$ has 4 intersections and the line $y=4$ has other 2 intersections$.
The equation $$ f(f(x)) = 0 $$ has thus 8 solutions.
