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== cocohearts is awesome ==

Revision as of 20:05, 1 June 2017

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 $2+4=\boxed{(D)6}$ 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]

See Also

2002 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

cocohearts is awesome

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