Difference between revisions of "2008 AMC 12A Problems/Problem 12"
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<math>\mathrm{(A)}\ [-1,1],[-1,0]\qquad\mathrm{(B)}\ [-1,1],[0,1]\qquad\textbf{(C)}\ [0,2],[-1,0]\qquad\mathrm{(D)}\ [1,3],[-1,0]\qquad\mathrm{(E)}\ [1,3],[0,1]</math> | <math>\mathrm{(A)}\ [-1,1],[-1,0]\qquad\mathrm{(B)}\ [-1,1],[0,1]\qquad\textbf{(C)}\ [0,2],[-1,0]\qquad\mathrm{(D)}\ [1,3],[-1,0]\qquad\mathrm{(E)}\ [1,3],[0,1]</math> | ||
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<math>g(x)</math> is defined if <math>f(x + 1)</math> is defined. Thus the domain is all <math>x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]</math>. | <math>g(x)</math> is defined if <math>f(x + 1)</math> is defined. Thus the domain is all <math>x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]</math>. | ||
Latest revision as of 05:31, 2 July 2024
Contents
Problem
A function has domain and range . (The notation denotes .) What are the domain and range, respectively, of the function defined by ?
Solution 1
is defined if is defined. Thus the domain is all .
Since , . Thus is the range of .
Thus the answer is .
Solution 2
Look at the vertical and horizontal translations, and we rewrite the function to a more recognizable to help us visualize.
Horizontal: There is one horizontal shift one unit to the left from the component making the domain
Vertical: There is one vertical mirror from the causing the range to become and then a vertical shift one unit upward from the causing the range to become .
This generates the answer of .
~PhysicsDolphin
See Also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.