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2008 AMC 12A Problems/Problem 12

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

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

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 $[ - 1,1],[0,1] \Rightarrow B$ .

See Also

2008 AMC 12A (Problems • 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

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Category: Introductory Algebra Problems

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