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> | ||
− | ==Solution== | + | ==Solution 1== |
<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>. | ||
Since <math>f(x + 1) \in [0,1]</math>, <math>- f(x + 1) \in [ - 1,0]</math>. Thus <math>g(x) = 1 - f(x + 1) \in [0,1]</math> is the range of <math>g(x)</math>. | Since <math>f(x + 1) \in [0,1]</math>, <math>- f(x + 1) \in [ - 1,0]</math>. Thus <math>g(x) = 1 - f(x + 1) \in [0,1]</math> is the range of <math>g(x)</math>. | ||
− | Thus the answer is <math>[ - 1,1],[0,1] \ | + | Thus the answer is <math>[- 1,1],[0,1] \longrightarrow \boxed{B}</math>. |
+ | |||
+ | ==Solution 2== | ||
+ | Look at the vertical and horizontal translations, and we rewrite the function to a more recognizable <math>-f(x+1) + 1</math> to help us visualize. | ||
+ | |||
+ | Horizontal: There is one horizontal shift one unit to the left from the <math>(x+1)</math> component making the domain <math>[-1, 1]</math> | ||
+ | |||
+ | Vertical: There is one vertical mirror from the <math>-f</math> causing the range to become <math>[-1, 0]</math> and then a vertical shift one unit upward from the <math>+ 1</math> causing the range to become <math>[0, 1]</math>. | ||
+ | |||
+ | This generates the answer of <math>\mathrm{(B)}</math>. | ||
+ | |||
+ | ~PhysicsDolphin | ||
==See Also== | ==See Also== |
Latest revision as of 06:31, 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.