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2002 AMC 12A Problems/Problem 19

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Problem

The graph of the function f is shown below. How many solutions does the equation f(f(x))=6 have?

size(300,300);defaultpen(fontsize(10pt)+linewidth(.8pt));dotfactor=4;pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6...

\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{6} solutions.

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See Also

2002 AMC 12A (ProblemsResources)
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
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
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