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

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A [[function]] <math>f</math> has [[domain]] <math>[0,2]</math> and [[range]] <math>[0,1]</math>. (The notation <math>[a,b]</math> denotes <math>\{x:a \le x \le b \}</math>.) What are the domain and range, respectively, of the function <math>g</math> defined by <math>g(x)=1-f(x+1)</math>?
 
A [[function]] <math>f</math> has [[domain]] <math>[0,2]</math> and [[range]] <math>[0,1]</math>. (The notation <math>[a,b]</math> denotes <math>\{x:a \le x \le b \}</math>.) What are the domain and range, respectively, of the function <math>g</math> defined by <math>g(x)=1-f(x+1)</math>?
  
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<math>\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]</math>
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<math>\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]</math>
  
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==Solution==
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==Solution 1==
 
<math>g(x)</math> is defined if <math>f(x + 1)</math> is defined. Thus the domain is all <math>x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]</math>.  
 
<math>g(x)</math> is defined if <math>f(x + 1)</math> is defined. Thus the domain is all <math>x| x + 1 \in [0,2] \rightarrow x \in [ - 1,1]</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>.  
 
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>.  
  
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Thus the answer is <math>[ - 1,1],[0,1] \Rightarrow B</math>.
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Thus the answer is <math>[- 1,1],[0,1] \longrightarrow \boxed{B}</math>.
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==Solution 2==
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Look at the vertical and horizontal translations, and we rewrite the function to a more recognizable <math>-f(x+1) + 1</math> to help us visualize.
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Horizontal: There is one horizontal shift one unit to the left from the <math>(x+1)</math> component making the domain <math>[-1, 1]</math>
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Vertical: There is one vertical mirror from the <math>-f</math> causing the range to become <math>[-1, 0]</math> and then a vertical shift one unit upward from the <math>+ 1</math> causing the range to become <math>[0, 1]</math>.
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This generates the answer of <math>\mathrm{(B)}</math>.
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~PhysicsDolphin
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 06:31, 2 July 2024

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 1

$g(x)$ 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 $\mathrm{(B)}$.

~PhysicsDolphin

See Also

2008 AMC 12A (Problems • Answer Key • 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
All AMC 12 Problems and Solutions

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