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

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Given an <math>x</math>, let <math>f(x)=t</math>. Obviously, to have <math>f(f(x))=6</math>, we need to have <math>f(t)=6</math>, and we already know when that happens. In other words, the solutions to <math>f(f(x))=6</math> are precisely the solutions to (<math>f(x)=-2</math> or <math>f(x)=1</math>).
 
Given an <math>x</math>, let <math>f(x)=t</math>. Obviously, to have <math>f(f(x))=6</math>, we need to have <math>f(t)=6</math>, and we already know when that happens. In other words, the solutions to <math>f(f(x))=6</math> are precisely the solutions to (<math>f(x)=-2</math> or <math>f(x)=1</math>).
  
Without actually computing the exact values, it is obvious from the graph that the equation <math>f(x)=-2</math> has two and <math>f(x)=1</math> has four different solutions, giving us a total of <math>2+4=(D)\boxed{6}</math> solutions.
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Without actually computing the exact values, it is obvious from the graph that the equation <math>f(x)=-2</math> has two and <math>f(x)=1</math> has four different solutions, giving us a total of <math>2+4=\boxed{(D)6}</math> solutions.
  
 
<asy>
 
<asy>

Revision as of 00:15, 2 July 2013

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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All AMC 12 Problems and Solutions
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