1982 AHSME Problems/Problem 19
Problem
Let 
for 
. The sum of the largest and smallest values of 
is
Solution
Note that at 
one of the three absolute values is equal to 
Without using absolute values, we rewrite as a piecewise function:
![\[f(x) = \begin{cases} (x-2)+(4-x)-(6-2x) & \mathrm{if} \ 2\leq x<3 \\ (x-2)+(4-x)-(2x-6) & \mathrm{if} \ 3\leq x<4 \\ (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 \end{cases},\]](http://latex.artofproblemsolving.com/a/b/4/ab4c849193927ab0c089903567108076ebd16b4c.png)
which simplifies to
![\[f(x) = \begin{cases} 2x-4 & \mathrm{if} \ 2\leq x<3 \\ -2x+8 & \mathrm{if} \ 3\leq x<4 \\ 0 & \mathrm{if} \ 4\leq x\leq8 \end{cases}.\]](http://latex.artofproblemsolving.com/0/2/1/021930b6003205f15bb7d7263a002ff9bf9b693f.png)
The graph of 
is shown below.
![[asy] /* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 10; int yMin = -2; int yMax = 4; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //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),E); label("$y$",(0,yMax),N); 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]](http://latex.artofproblemsolving.com/8/3/7/83733cfd123f55c48f1b588b500435a9fb53ab16.png)
The largest value of is 
and the smallest value of is So, their sum is 
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
