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

Revision as of 00:21, 4 February 2019

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 9: / Line 9: @@
 Since <math>f(x + 1) \in [0,1]</math>, <math>- f(x + 1) \in [ - 1,0]</math>. Thus <math>g(x) = 1 - f(x + 1) \in [0,1]</math> is the range of <math>g(x)</math>.
-Thus the answer is <math>[ - 1,1],[0,1] \longrightarrow \boxed{B}</math>.
+Thus the answer is <math>[- 1,1],[0,1] \longrightarrow \boxed{B}</math>.
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Difference between revisions of "2008 AMC 12A Problems/Problem 12"

Revision as of 00:21, 4 February 2019

Problem

Solution

See Also