1980 AHSME Problems/Problem 22

Problem

For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1, x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is

$\text{(A)} \ \frac{1}{3} \qquad \text{(B)} \ \frac{1}{2} \qquad \text{(C)} \ \frac{2}{3} \qquad \text{(D)} \ \frac{5}{2} \qquad \text{(E)}\ \frac{8}{3}$

Solution

The first two given functions intersect at $\left(\frac{1}{3},\frac{7}{3}\right)$, and last two at $\left(\frac{2}{3},\frac{8}{3}\right)$. Therefore $$f(x)=\left\{ \begin{matrix} 4x+1 & x<\frac{1}{3} \\ x+2 & \frac{1}{3}>x>\frac{2}{3} \\ -2x+4 & x>\frac{2}{3} \end{matrix}\right.$$ Which attains a maximum at $\boxed{(E)\ \frac{8}{3}}$

See also

 1980 AHSME (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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

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