Difference between revisions of "2008 AMC 12A Problems/Problem 12"
Miller4math (talk | contribs) m (→Solution) |
(→Solution) |
||
| (5 intermediate revisions by 4 users not shown) | |||
| Line 1: | Line 1: | ||
==Problem == | ==Problem == | ||
| − | A function <math>f</math> has domain <math>[0,2]</math> and range <math>[0,1]</math>. (The notation <math>[a,b]</math> denotes <math>\{x:a \le x \le b \}</math>.) What are the domain and range, respectively, of the function <math>g</math> defined by <math>g(x)=1-f(x+1)</math>? | + | A [[function]] <math>f</math> has [[domain]] <math>[0,2]</math> and [[range]] <math>[0,1]</math>. (The notation <math>[a,b]</math> denotes <math>\{x:a \le x \le b \}</math>.) What are the domain and range, respectively, of the function <math>g</math> defined by <math>g(x)=1-f(x+1)</math>? |
| − | <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== | ||
| Line 9: | Line 9: | ||
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>. |
==See Also== | ==See Also== | ||
{{AMC12 box|year=2008|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2008|ab=A|num-b=11|num-a=13}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 01:21, 4 February 2019
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
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)
.
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. 
