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

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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>.  
 
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>.
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Thus the answer is <math>[- 1,1],[0,1] \longrightarrow \boxed{B}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 01:21, 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 $[- 1,1],[0,1] \longrightarrow \boxed{B}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
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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