Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
m (→‎Solution)
 
(6 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
<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>
 
<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>
  
−
==Solution==
+
==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>.  
  
−
Thus the answer is <math>[ - 1,1],[0,1] \longightarrow \boxed{B}</math>.
+
Thus the answer is <math>[- 1,1],[0,1] \longrightarrow \boxed{B}</math>.
 +
 
 +
==Solution 2==
 +
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.
 +
 
 +
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>
 +
 
 +
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>.
 +
 
 +
This generates the answer of <math>\mathrm{(B)}</math>.
 +
 
 +
~PhysicsDolphin
  
 
==See Also==
 
==See Also==

Latest revision as of 05:31, 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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12A_Problems/Problem_12&oldid=221497"