Question

The maximum value of $f\left(x\right)=\frac{x}{\left[4+x+x^2\right]}$ on $\left[-1,1\right]$ is

• A. $-\left(\frac{1}{3}\right)$
• B. $-\left(\frac{1}{4}\right)$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

The correct option is D

$\frac{1}{6}$

Explanation for the correct option:

Step $1:$ Solve for the maximum value of $f\left(x\right)=\frac{x}{\left[4+x+x^2\right]}$ on $\left[-1,1\right]$

$$\begin{gathered} f\left(x\right)=\frac{x}{\left[4+x+x^2\right]} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1(4+x+x^2)-x(1+2x)}{{\left[4+x+x^2\right]}^2}\left[\because \frac{d}{dx}\left[\frac{f\left(x\right)}{g\left(x\right)}\right]=\frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{{\left[g\left(x\right)\right]}^2}\right] \end{gathered}$$

For a local maxima or a local minima, we must have

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

$f\text{'}\left(x\right)>0,\forall x\in (-2,2)$

$f\left(x\right)$ is increasing in $x\in (-2,2)$

The values of $f\left(x\right)$ at extreme points are given by

$$\begin{gathered} f(-1)=\frac{-1}{4-1+{(-1)}^2}=\frac{-1}{4} \\ f\left(1\right)=\frac{1}{4+1+{\left(1\right)}^2}=\frac{1}{6} \end{gathered}$$

$f\left(x\right)$ is increasing in $\left[-1,1\right]$ and therefore the maximum value of $f\left(x\right)$ occurs at $x=1$

Hence the maximum value of $f\left(x\right)=\frac{x}{\left[4+x+x^2\right]}$ on $\left[-1,1\right]$ is $\frac{1}{6}.$

Hence, option(D) is correct answer.