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The maximum value

Question:

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

(a) $-\frac{1}{4}$

(b) $-\frac{1}{3}$

(C) $\frac{1}{6}$

(d) $\frac{1}{5}$

Solution:

(C) $\frac{1}{6}$

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

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

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

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

$\Rightarrow \frac{4+x+x^{2}-x(1+2 x)}{\left(4+x+x^{2}\right)^{2}}=0$

$\Rightarrow 4+x+x^{2}-x(1+2 x)=0$

$\Rightarrow 4-x^{2}=0$

$\Rightarrow x=\pm 2 \notin[-1,1]$

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

$f(1)=\frac{1}{4+1+1^{2}}=\frac{1}{6}$

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

Thus, $\frac{1}{6}$ is the maximum value.

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