pancha3

2021-08-13

$f\left(x\right)=\frac{x}{{x}^{2}+1}$
Find the absolute extreme values on [-7,7]

First direction:
$f\left(x\right)=\frac{x}{{x}^{2}+1}$ on [-7,7]
${f}^{\prime }\left(x\right)=\frac{d}{dx}\left(\frac{x}{{x}^{2}+1}\right)$
$=\frac{\left({x}^{2}+1\right)-x\left(2x\right)}{{\left({x}^{2}+1\right)}^{2}}$
$=\frac{1-{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}$
Equate above first derivative to zero to obtain the extreme points:
${f}^{\prime }\left(x\right)=0$
$\frac{1-{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}=0$
$1-{x}^{2}=0$
${x}^{2}=1$
$x=±1$
Compute the second derivative:
${f}^{\prime }\left(x\right)=\frac{1-{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}$
$f{}^{″}\left(x\right)=\frac{d}{dx}\left(\frac{1-{x}^{2}}{{\left({x}^{2}+1\right)}^{2}}\right)$
$=\frac{{\left({x}^{2}+1\right)}^{2}\left(-2x\right)-\left(1-{x}^{2}\right)\left(4x\left({x}^{2}+1\right)\right)}{{\left({x}^{2}+1\right)}^{4}}$
$f{}^{″}\left(1\right)=\frac{1\left(-2\right)-0}{1}=-2<0$ [max at 1]
$f{}^{″}\left(-1\right)=\frac{1\left(2\right)-0}{1}=2<0$ [min at -1]
Thus,
Max value is $f\left(1\right)=\underset{of1}{=}\left(\frac{x}{{x}^{2}+1}\right)=\frac{1}{2}$
Min value is $f\left(-1\right)=\underset{of-1}{=}\left(\frac{x}{{x}^{2}+1}\right)=-\frac{1}{2}$

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