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From the graph we can find that, $f(-5)=f(-1)=f(9)=0$

enter image description here

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If we restrict ourselves just on this graph, I mean we accept no other graph exists outer than $(-10,13)$. Then you need just to find the solution of $f(x)=4$, which is 4 distinct solutions $x_1,x_2,x_3,x_4$.

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Also $f(x)=12$ gives $f(f(x))=15$. – Juris Feb 1 '13 at 14:30
@Juris: Yes!. It 's omitted neglectfully and so incorrectly. Thanks for remarking that. – Babak S. Feb 1 '13 at 14:41
@Juris: Please edit your deleted answer and add what you had noted me in above comment as an completeing point. – Babak S. Feb 2 '13 at 6:38

As Babak points out, we assume the domain of our function is restricted to the interval $(-10, 13)$:

  • When $f(x) =4,\;$ $f(f(4)) = 15:\;$

    Find the x-values that correspond to all the points where $f(x) = 4$: you should find four such $x$-values: $x = -8, 1, 7.5, 10$.

  • When $f(x) = 12$, note that $f(f(12)) = 15:$

    Find the x-values that correspond to where $f(x) = 12$. Again, there should be three or four such values, depending on whether you include the value at the far, far left of the graph: (approximations by "eying up" $x =? = -10, x = 3.5, 5.25, 11.5$

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We are to find the solutions for $f(f(x)) = 15$.

From the graph, $f(4) = 15$ and $f(12) = 15$.

The required solutions will be those values of $x$ for which $f(x) = 4$ and $f(x) = 12$.

From the graph, the value of function $f(x) = 4$ at four different values of $x$, namely $x = –8, 1, 7.5, 10$.

The value of the function $f(x) = 12$ at three different points, namely $x = 3.5, 5.25, 11.5$.

Hence, the given equation has 7 solutions.

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