Difference between revisions of "1982 AHSME Problems/Problem 19"
MRENTHUSIASM (talk | contribs) (→Solution) |
MRENTHUSIASM (talk | contribs) m (→Solution) |
||
Line 17: | Line 17: | ||
(x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 | (x-2)+(x-4)-(2x-6) & \mathrm{if} \ 4\leq x\leq8 | ||
\end{cases},</cmath> | \end{cases},</cmath> | ||
− | which | + | which simplifies to |
<cmath>f(x) = \begin{cases} | <cmath>f(x) = \begin{cases} | ||
2x-4 & \mathrm{if} \ 2\leq x<3 \\ | 2x-4 & \mathrm{if} \ 2\leq x<3 \\ |
Revision as of 20:19, 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: which simplifies to The graph of is shown below. 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.