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2002 AMC 12A Problems/Problem 19: Difference between revisions

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New page: == Problem == The graph of the function <math>f</math> is shown below. How many solutions does the equation <math>f(f(x))=6</math> have? <asy> size(300,300); defaultpen(fontsize(10pt)+l...
 
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<asy>
<asy>
size(300,300);
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=4;
dotfactor=4;
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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=\boxed{6}</math> solutions.
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>
size(300,300);
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=4;
dotfactor=4;
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yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6));
yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6));
</asy>
</asy>
== Video Solution ==
https://youtu.be/d9A-UTh07Rc


== See Also ==
== See Also ==
{{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}


{{AMC12 box|year=2002|ab=A|num-b=18|num-a=20}}
[[Category:Introductory Algebra Problems]]
 
 
{{MAA Notice}}

Latest revision as of 10:09, 18 July 2023

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]

Video Solution

https://youtu.be/d9A-UTh07Rc

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


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