Question

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

Answer 1

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

The function f(x) would attain its maximum value, when the value of 4x² + 2x + 1 is minimum.

Let g(x) = 4x² + 2x + 1

$\therefore g\text{'}\left(x\right)=8x+2$

For maxima or minima,

$g\text{'}\left(x\right)=0$

$\Rightarrow 8x+2=0$

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

Now,

$g\text{'}\text{'}\left(x\right)=8>0$

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

Minimum value of function g(x)

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

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

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

$=\frac{3}{4}$

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

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


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