# Consider the following function. f(x)=frac{x^{2}}{x^{2}-81} a) To find the critucal numbers of f. b) To find the open interval on which function is increasing or decreasing. c) To identify the relative extremum.

Consider the following function. $f\left(x\right)=\frac{{x}^{2}}{{x}^{2}-81}$ a) To find the critucal numbers of f. b) To find the open interval on which function is increasing or decreasing. c) To identify the relative extremum.
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SoosteethicU

a) Let us find first derivative of the function f. ${f}^{\prime }\left(x\right)=\frac{\left({x}^{2}-81\right)2x-{x}^{2}\left(2x\right)}{{\left({x}^{2}-81\right)}^{2}}=\frac{2{x}^{3}-162x-2{x}^{3}}{{\left({x}^{2}-81\right)}^{2}}=\frac{-162x}{{\left({x}^{2}-81\right)}^{2}}$ We know that critical numbers are those number where the derivative vanishes or does not exist. ${f}^{\prime }\left(x\right)=0⇒x=0$ Clearly ${f}^{\prime }\left(x\right)$ does not exist when ${x}^{2}-81=0⇒x=±9$ Thus the critical numbers are - $x=0,-9,9$ b) For increasing, ${f}^{\prime }\left(x\right)>0$
$⇒\frac{-162x}{{\left({x}^{2}-81\right)}^{2}}>0$
$⇒-162x>0$
$⇒x<0$
$\therefore x\in \left(-\mathrm{\infty },0\right)$ b) For decreasing ${f}^{\prime }\left(x\right)<0$
$⇒\frac{-162x}{\left({x}^{2}-81{\right)}^{2}}<0$$⇒-162x<0$
$⇒x>0$
$\therefore x\in \left(0,\mathrm{\infty }\right)$ Thus, increasing on $\left(-\mathrm{\infty },0\right)$ decreasing on $\left(0,\mathrm{\infty }\right)$ c) By first derivative test,

$\therefore x=0$ gives relative maximum. $y=0$ and relative minimum does not exist. Therefore, relative maximum $\left(x,y\right)=\left(0,0\right)$ relative minimum $\left(x,y\right)=DNE$