How do you find the local maximum and minimum values of $f(x) = x / (x^2 + 81)$ using both the First and Second Derivative Tests?

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Answer 1 · 2018-01-02T09:50:24

Minimum $(-9,-1/18)$ , Maximum $(9,1/18)$

$f(x)=x/(x^2+81)$

$D_f=RR$

$(-x^2+81)/(x^2+81)^2$ $=$

$-(x^2-81)/(x^2+81)^2$

$f'(x)=0$ $<=> x^2-81=0 <=> x^2=81 <=> (x=9 or x=-9)$

Let's plug in $f'$ a value , $x_1in$ $(-9,9)$ . For example, for $x_1=0$
we get

$f'(0)=-(-81)/81^2=81/81^2>0$

Let's plug in $f'$ a value which is $>9$ . For example, for $x_2=10$
we get

$f'(10)=-(100-81)/(100+81)^2=-19/181^2<0$

Let's plug in $f'$ a value which is $<-9$ . For example, for $x_3=-10$
we get

$f'(-10)=-(100-81)/(100+81)^2=-19/181^2<0$

As a result we have:

$f$ continuous in $(-oo,-9]$ and $f'(x)<0$ for $x$ $in$ $(-oo,-9)$
so $f$ is strictly decreasing in $(-oo,-9]$
$f$ continuous in $[-9,9]$ and $f'(x)>0$ for $x$ $in$ $(-9,-9)$
so $f$ is strictly increasing in $[-9,9]$
$f$ continuous in $[9,+oo)$ and $f'(x)<0$ for $x$ $in$ $(9,+oo)$
so $f$ is strictly decreasing in $[9,+oo)$

$f$ is decreasing in $(-oo,-9]$ and increasing in $[-9,9]$
therefore $f$ has a local minimum at $x=-9$

$f$ is increasing in $[-9,9]$ and decreasing in $[9,+oo)$
therefore $f$ has a local maximum at $x=9$

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How do you find the local maximum and minimum values of f(x) = x / (x^2 + 81)f(x) = x / (x^2 + 81) using both the First and Second Derivative Tests?