Difference between revisions of "2008 AMC 12A Problems/Problem 12"
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==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>\textbf{(A)}\ [ - 1,1],[ - 1,0] \qquad \textbf{(B)}\ [ - 1,1],[0,1] \qquad \textbf{(C)}\ [0,2],[ - 1,0] \qquad \textbf{(D)}\ [1,3],[ - 1,0] \qquad \textbf{(E)}\ [1,3],[0,1]</math> | <math>\textbf{(A)}\ [ - 1,1],[ - 1,0] \qquad \textbf{(B)}\ [ - 1,1],[0,1] \qquad \textbf{(C)}\ [0,2],[ - 1,0] \qquad \textbf{(D)}\ [1,3],[ - 1,0] \qquad \textbf{(E)}\ [1,3],[0,1]</math> | ||
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==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]] | ||
Revision as of 17:06, 20 February 2008
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 
?
![$\textbf{(A)}\ [ - 1,1],[ - 1,0] \qquad \textbf{(B)}\ [ - 1,1],[0,1] \qquad \textbf{(C)}\ [0,2],[ - 1,0] \qquad \textbf{(D)}\ [1,3],[ - 1,0] \qquad \textbf{(E)}\ [1,3],[0,1]$](http://latex.artofproblemsolving.com/9/c/3/9c34c533e9603d76b4950aabba8383821c18cad2.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 ![$[ - 1,1],[0,1] \Rightarrow B$](http://latex.artofproblemsolving.com/b/b/d/bbdbdbdcce7ef8f16560ce957da2e90999828946.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 | |
