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Difference between revisions of "1982 AHSME Problems/Problem 19"

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label("$(4,0)$",A[2],(0,-1.5),UnFill);
 
label("$(4,0)$",A[2],(0,-1.5),UnFill);
 
label("$(8,0)$",A[3],(0,-1.5),UnFill);
 
label("$(8,0)$",A[3],(0,-1.5),UnFill);
 +
 +
label("$\phantom{xxx}$",(10,0),(0.5,-1.5),UnFill);
 +
label("$\phantom{YYY}$",(0,4),(-2,0.5),UnFill);
 +
label("$x$",(10,0),(1,0));
 +
label("$y$",(0,4),N);
 
</asy>
 
</asy>
 
The largest value of <math>f(x)</math> is <math>2,</math> and the smallest value of <math>f(x)</math> is <math>0.</math> So, their sum is <math>\boxed{\textbf {(B)}\ 2}.</math>
 
The largest value of <math>f(x)</math> is <math>2,</math> and the smallest value of <math>f(x)</math> is <math>0.</math> So, their sum is <math>\boxed{\textbf {(B)}\ 2}.</math>

Revision as of 03:26, 13 September 2021

Problem

Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is

$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$

Solution

Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$

Without using absolute values, we rewrite $f(x)$ as a piecewise function: \[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\] which simplifies to \[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\] The graph of $y=f(x)$ is shown below. [asy] /* Made by MRENTHUSIASM */ size(220); import TrigMacros; rr_cartesian_axes(-2,10,-2,4,useticks=true); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(2,0)$",A[0],(0,-1.5),UnFill); label("$(3,2)$",A[1],(0,1.5),UnFill); label("$(4,0)$",A[2],(0,-1.5),UnFill); label("$(8,0)$",A[3],(0,-1.5),UnFill); label("$\phantom{xxx}$",(10,0),(0.5,-1.5),UnFill); label("$\phantom{YYY}$",(0,4),(-2,0.5),UnFill); label("$x$",(10,0),(1,0)); label("$y$",(0,4),N); [/asy] The largest value of $f(x)$ is $2,$ and the smallest value of $f(x)$ is $0.$ So, their sum is $\boxed{\textbf {(B)}\ 2}.$

~MRENTHUSIASM

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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