# Difference between revisions of "2002 AMC 12A Problems/Problem 19"

## Problem

The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x))=6$ have? $[asy] size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; draw(P1--P2--P3--P4--P5); dot("(-7, -4)",P1); dot("(-2, 6)",P2,LeftSide); dot("(1, 6)",P4); dot("(5, -6)",P5); xaxis("x",-7.5,7,Ticks(xticks),EndArrow(6)); yaxis("y",-6.5,7,Ticks(yticks),EndArrow(6)); [/asy]$ $\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }6 \qquad \text{(E) }7$

## Solution

First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$.

Given an $x$, let $f(x)=t$. Obviously, to have $f(f(x))=6$, we need to have $f(t)=6$, and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ( $f(x)=-2$ or $f(x)=1$).

Without actually computing the exact values, it is obvious from the graph that the equation $f(x)=-2$ has two and $f(x)=1$ has four different solutions, giving us a total of $2+4=(D)\boxed{6}$ solutions. $[asy] size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; path graph = P1--P2--P3--P4--P5; path line1 = (-7,1)--(6,1); path line2 = (-7,-2)--(6,-2); draw(graph); draw(line1, red); draw(line2, red); dot("(-7, -4)",P1); dot("(-2, 6)",P2,LeftSide); dot("(1, 6)",P4); dot("(5, -6)",P5); dot(intersectionpoints(graph,line1),red); dot(intersectionpoints(graph,line2),red); xaxis("x",-7.5,7,Ticks(xticks),EndArrow(6)); yaxis("y",-6.5,7,Ticks(yticks),EndArrow(6)); [/asy]$