waigaK

2020-10-18

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.

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 $\left(x,y\right)=DNE$

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