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Solution:

The given function is $f(x)=\frac{1}{4 x^{2}+2 x+1}$.

The function $f(x)$ would attain its maximum value, when the value of $4 x^{2}+2 x+1$ is minimum.

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

$\therefore g^{\prime}(x)=8 x+2$

For maxima or minima,

$g^{\prime}(x)=0$

$\Rightarrow 8 x+2=0$

 

$\Rightarrow x=-\frac{1}{4}$

Now,

$g^{\prime \prime}(x)=8>0$

So, $x=-\frac{1}{4}$ is the point of local minimum of $g(x)$

Minimum value of function g(x)

$=g\left(-\frac{1}{4}\right)$

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

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

 

$=\frac{3}{4}$

$\therefore$ Maximum value of $f(x)=\frac{1}{\text { Minimum value of } g(x)}=\frac{1}{\left(\frac{3}{4}\right)}=\frac{4}{3}$

Thus, the maximum value of the function $f(x)=\frac{1}{4 x^{2}+2 x+1}$ is $\frac{4}{3}$.

If $f(x)=\frac{1}{4 x^{2}+2 x+1}$, then its maximum value is $\frac{4}{3}$