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== | ||
| Line 15: | Line 26: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 05:31, 2 July 2024
Contents
Problem
A function 
has domain ![$[0,2]$](http://latex.artofproblemsolving.com/9/0/7/9076ffb2b9c051f1b6f07bf18066581d57216258.png)
and range ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
. (The notation ![$[a,b]$](http://latex.artofproblemsolving.com/8/e/c/8ecbd1ba3da8f2adef66a63f2ab32c47e63fa734.png)
denotes 
.) What are the domain and range, respectively, of the function 
defined by 
?
![$\mathrm{(A)}\ [-1,1],[-1,0]\qquad\mathrm{(B)}\ [-1,1],[0,1]\qquad\textbf{(C)}\ [0,2],[-1,0]\qquad\mathrm{(D)}\ [1,3],[-1,0]\qquad\mathrm{(E)}\ [1,3],[0,1]$](http://latex.artofproblemsolving.com/7/8/8/78886ac42b5d0e223a915fbc087f11ad5bb80640.png)
Solution 1
is defined if 
is defined. Thus the domain is all ![$x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$](http://latex.artofproblemsolving.com/d/8/b/d8b6624181948180d27595212f0b42c0aceed048.png)
.
Since ![$f(x + 1) \in [0,1]$](http://latex.artofproblemsolving.com/6/2/9/629de8878872a37847132bd922824fc8b712810d.png)
, ![$- f(x + 1) \in [ - 1,0]$](http://latex.artofproblemsolving.com/c/c/9/cc930a8e58693cb8571f0405363867f1f5d76983.png)
. Thus ![$g(x) = 1 - f(x + 1) \in [0,1]$](http://latex.artofproblemsolving.com/8/8/3/8834c52dfda63c5f6ffb6a790b72dadf548aa29b.png)
is the range of .
Thus the answer is ![$[- 1,1],[0,1] \longrightarrow \boxed{B}$](http://latex.artofproblemsolving.com/1/f/0/1f0db379b8d21a2b1747508605654327f5c0a7e9.png)
.
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 ![$[-1, 1]$](http://latex.artofproblemsolving.com/5/c/3/5c3818b9565a33fd3aadba10026d32c5e3eea90f.png)
Vertical: There is one vertical mirror from the 
causing the range to become ![$[-1, 0]$](http://latex.artofproblemsolving.com/1/c/5/1c551e8202d12d189d2ddc127ffbbc8cb2cf00cc.png)
and then a vertical shift one unit upward from the 
causing the range to become ![$[0, 1]$](http://latex.artofproblemsolving.com/a/b/1/ab178d831a786b92cb4c9ddc2d33578223036f98.png)
.
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. 
