1982 AHSME Problems/Problem 19
Revision as of 19:12, 12 September 2021 by 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)...")
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.
DIAGRAM NEEDED
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 | |
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| All AHSME Problems and Solutions | ||
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