Difference between revisions of "1982 AHSME Problems/Problem 19"
MRENTHUSIASM (talk | contribs) (Created page with "== Problem == Let <math>f(x)=|x-2|+|x-4|-|2x-6|</math> for <math>2 \leq x\leq 8</math>. The sum of the largest and smallest values of <math>f(x)</math> is <math>\textbf {(A)...") |
MRENTHUSIASM (talk | contribs) (→Solution) |
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\end{cases}.</cmath> | \end{cases}.</cmath> | ||
The graph of <math>y=f(x)</math> is shown below. | The graph of <math>y=f(x)</math> is shown below. | ||
| + | <asy> | ||
| + | /* Made by MRENTHUSIASM */ | ||
| + | size(220); | ||
| + | import TrigMacros; | ||
| − | + | rr_cartesian_axes(-2,10,-2,4,useticks=true); | |
| + | pair A[]; | ||
| + | A[0] = (0,0); | ||
| + | A[1] = (2,0); | ||
| + | A[2] = (3,2); | ||
| + | A[3] = (4,0); | ||
| + | A[4] = (8,0); | ||
| + | |||
| + | draw(A[0]--A[1]--A[2]--A[3]--A[4],red+linewidth(1.5)); | ||
| + | |||
| + | for(int i = 0; i <= 4; ++i) { | ||
| + | dot(A[i],red+linewidth(4.5)); | ||
| + | } | ||
| + | |||
| + | label("$(0,0)$",A[0],(0,-1.5),UnFill); | ||
| + | label("$(2,0)$",A[1],(0,-1.5),UnFill); | ||
| + | label("$(3,2)$",A[2],(0,1.5),UnFill); | ||
| + | label("$(4,0)$",A[3],(0,-1.5),UnFill); | ||
| + | label("$(8,0)$",A[4],(0,-1.5),UnFill); | ||
| + | </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 20:17, 12 September 2021
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 simplify 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(220); import TrigMacros; rr_cartesian_axes(-2,10,-2,4,useticks=true); pair A[]; A[0] = (0,0); A[1] = (2,0); A[2] = (3,2); A[3] = (4,0); A[4] = (8,0); draw(A[0]--A[1]--A[2]--A[3]--A[4],red+linewidth(1.5)); for(int i = 0; i <= 4; ++i) { dot(A[i],red+linewidth(4.5)); } label("$(0,0)$",A[0],(0,-1.5),UnFill); label("$(2,0)$",A[1],(0,-1.5),UnFill); label("$(3,2)$",A[2],(0,1.5),UnFill); label("$(4,0)$",A[3],(0,-1.5),UnFill); label("$(8,0)$",A[4],(0,-1.5),UnFill); [/asy]](http://latex.artofproblemsolving.com/f/7/a/f7acc3dc3d120687b500519a328adbd912b59890.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. 
