The maximum value of \( f(x) = \frac{1}{4x^2 + 2x + 1} \) is

\( \frac{4}{3} \)

The correct answer is Option (2) → \( \frac{4}{3} \)

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

Step 1: Complete the square in the denominator:

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

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

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

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

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

So:

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

Minimum value of denominator: occurs when $\left(x + \frac{1}{4}\right)^2 = 0$

Minimum = $\frac{3}{4}$

Maximum of $f(x)$: is $\frac{1}{\frac{3}{4}} = \frac{4}{3}$