Question

Let $f(x) = \max(2x + 1, 3 - 4x)$, where $x$ is any real number. Then the minimum possible value of $f(x)$ is:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{4}{3}$
• E. $\frac{5}{3}$

Hint:

For minimization of $\max(f,g)$, solve $f=g$ to balance both expressions.

Answer 1

The Correct Option is D

For $\max(2x+1, 3-4x)$ to be minimum, set $2x + 1 = 3 - 4x$: $6x = 2 \Rightarrow x = \frac{1}{3}$, value = $2 \cdot \frac{1}{3} + 1 = \frac{5}{3}$? Wait, check: $2x+1 = 2/3 + 1 = 5/3$? That’s option (5)? But $\max$ minimization occurs when both are equal. On recalculation, $2(1/3)+1 = 5/3$ indeed, so answer is (5) not (4). \[ \boxed{\frac{5}{3}} \]