2008 AMC 12A Problems/Problem 12

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 $g(x)=1-f(x+1)$ ?

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

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
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2008 AMC 12A Problems/Problem 12

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