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

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

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 $[- 1,1],[0,1] \longrightarrow \boxed{B}$ .

Solution 2

Look at the vertical and horizontal translations, and we rewrite the function to a more recognizable $-f(x+1) + 1$ to help us visualize.

Horizontal: There is one horizontal shift one unit to the left from the $(x+1)$ component making the domain $[-1, 1]$

Vertical: There is one vertical mirror from the $-f$ causing the range to become $[-1, 0]$ and then a vertical shift one unit upward from the $+ 1$ causing the range to become $[0, 1]$ .

This generates the answer of $boxed{B}$ .

~PhysicsDolphin

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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2008 AMC 12A Problems/Problem 12

Contents

Problem

Solution

Solution 2

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