# 1982 AHSME Problems/Problem 19

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## 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. $[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("x",(xMax,0),(2,0)); label("y",(0,yMax),(0,2)); pair A[]; A[0] = (2,0); A[1] = (3,2); A[2] = (4,0); A[3] = (8,0); draw(A[0]--A[1]--A[2]--A[3],red+linewidth(1.5)); for(int i = 0; i <= 3; ++i) { dot(A[i],red+linewidth(4.5)); } label("(2,0)",A[0],(0,-1.5),UnFill); label("(3,2)",A[1],(0,1.5),UnFill); label("(4,0)",A[2],(0,-1.5),UnFill); label("(8,0)",A[3],(0,-1.5),UnFill); [/asy]$ 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