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Q.
Let $f(x)=\operatorname{Min} .(4 x+1, x+2,-2 x+4)$. Then the maximum value of $f(x)$ is
Relations and Functions - Part 2
Solution:
The three lines intersect at $x=\frac{1}{3}, \frac{1}{2}, \frac{2}{3}$.
In the interval $-\infty< x \leq \frac{1}{3}, 4 x+1$ 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 $\frac{1}{3} \leq x \leq \frac{2}{3}, x+2$ is minimum and its maximum value is $\frac{2}{3}+2=\frac{8}{3}$. In the interval $\frac{2}{3} \leq x< \infty,-2 x+4$ is minimum and its maximum is also $\frac{8}{3}$ at $x=\frac{2}{3}$.
Therefore, the answer is $\frac{8}{3}$