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

$\frac{4}{3}$

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

f(x) → Maximum when g(x) = $4x^2+2x+ 1$ is minimum 

so $g(x) = 4x^2+2x+ 1$

differentiating g(x) wrt x

g'(x) = 8x + 2

for g'(x) = 0  ⇒ 8x + 2 = 0

$x = \frac{-2}{8} = \frac{-1}{4}$

differentiating g'(x) wrt x

so g''(x) = 8

$g''(\frac{-1}{4}) = 8 > 0$  ⇒  point of local minima

⇒  f(x) maximum at $x = \frac{-1}{4}$

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

$\frac{1}{\frac{1}{4}-\frac{2}{4}+1}$

$=\frac{4}{3}$