If $f(x) = \frac{1}{4x^2 + 2x + 1}; x \in \mathbb{R}$, then find the maximum value of $f(x)$.

$\frac{4}{3}$

The correct answer is Option (3) → $\frac{4}{3}$ ##

$f(x) = \frac{1}{4x^2 + 2x + 1}$

Let $g(x) = 4x^2 + 2x + 1$

$= 4\left(x^2 + 2x \cdot \frac{1}{4} + \frac{1}{16}\right) + \frac{3}{4}$

$= 4\left(x + \frac{1}{4}\right)^2 + \frac{3}{4}$

$∴$ Maximum value of $f(x) = \frac{1}{3/4} = \frac{4}{3}$