Difference between revisions of "2008 AMC 12A Problems/Problem 12"

Revision as of 01:19, 26 April 2008

Problem

A function $f$ has domain $[0,2]$ and range $[0,1]$ . (The notation $[a,b]$ denotes $\{x:a \le x \le b \}$ .) What are the domain and range, respectively, of the function $g$ defined by $g(x)=1-f(x+1)$ ?

$\mathrm{(A)}\ [-1,1],[-1,0]\qquad\mathrm{(B)}\ [-1,1],[0,1]\qquad\textbf{(C)}\ [0,2],[-1,0]\qquad\mathrm{(D)}\ [1,3],[-1,0]\qquad\mathrm{(E)}\ [1,3],[0,1]$

Solution

$g(x)$ is defined if $f(x + 1)$ is defined. Thus the domain is all $x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]$ .

Since $f(x + 1) \in [0,1]$ , $- f(x + 1) \in [ - 1,0]$ . Thus $g(x) = 1 - f(x + 1) \in [0,1]$ is the range of $g(x)$ .

Thus the answer is $[ - 1,1],[0,1] \Rightarrow B$ .

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

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 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>\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==

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Difference between revisions of "2008 AMC 12A Problems/Problem 12"

Revision as of 01:19, 26 April 2008

Problem

Solution

See Also