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=\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> | ||
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yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6)); | yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6)); | ||
</asy> | </asy> | ||
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| + | == Video Solution == | ||
| + | https://youtu.be/d9A-UTh07Rc | ||
== See Also == | == See Also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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| + | {{MAA Notice}} | ||
Latest revision as of 10:09, 18 July 2023
Contents
Problem
The graph of the function 
is shown below. How many solutions does the equation 
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]](http://latex.artofproblemsolving.com/d/8/e/d8ea42d74287fb95665640e991c92aa37ba2c70f.png)
Solution
First of all, note that the equation 
has two solutions: 
and 
.
Given an 
, let 
. Obviously, to have , we need to have , and we already know when that happens. In other words, the solutions to are precisely the solutions to (
or 
).
Without actually computing the exact values, it is obvious from the graph that the equation has two and has four different solutions, giving us a total of 
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]](http://latex.artofproblemsolving.com/6/5/3/653468fb472ce591f7429c348a1f2b68deca3c12.png)
Video Solution
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. 
