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) | + | 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 is shown below. How many solutions does the equation have?
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.
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 |