Question

Let $f(x)=\mathrm{Min}.(4x+1,x+2,-2x+4)$. Then the maximum value of $f(x)$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{8}{3}$

Answer 1

Answer: (D)

The three lines intersect at $x=\frac{1}{3},\frac{1}{2},\frac{2}{3}$.
In the interval $-\infty <x\le \frac{1}{3},4x+1$ is minimum. Thus the maixmum in this interval is $4\left(\frac{1}{3}\right)+1=\frac{7}{3}$.
In the same manner in the interval $\frac{1}{3}\le x\le \frac{2}{3},x+2$ is minimum and its maximum value is $\frac{2}{3}+2=\frac{8}{3}$. In the interval $\frac{2}{3}\le x<\infty ,-2x+4$ is minimum and its maximum is also $\frac{8}{3}$ at $x=\frac{2}{3}$.
Therefore, the answer is $\frac{8}{3}$