Question

How do you find the local max and min for `f(x) = x / (x^2 + 81)`?

Answer 1

`f(9)=1/18" is local Maxima."`

`f(-9)=-1/18" is local Minima."`

Explanation:

We know that `f` has a local maxima /minima at `x=a`, then,

#(1) f'(a)=0, and, (2) f''(a) lt or gt 0" according as maxima or minima, resp."#

Now, `f(x)=x/(x^2+81) :. f'(x)=((x^2+81)(1)-x(2x))/(x^2+81)^2, i.e.,`

`f'(x)=(81-x^2)/(x^2+81)^2`.

`:. f'(x)=0 rArr x=+-9`

Further, `f''(x)={(x^2+81)^2(-2x)-(81-x^2)2(x^2+81)2x}/(x^2+81)^4`

`={(x^2+81)(-2x)(x^2+81+2(81-x^2))}/(x^2+81)^4`

`={-2x(3(81)-x^2)}/(x^2+81)^3`

Now, `f''(9)={-2(9)(3(81)-81)}/(81+81)^3={-2(9)(2(81))}/(2^3*81^3) lt 0`

`rArr f(9)=9/(81+81)=9/(2*81)=1/18" is local Maxima."`

Finally, `f''(-9)={-2(-9)(2(81))}/(2^3*81^3) gt 0`

`rArr f(-9)=-9/(2*81)=-1/18" is local Minima."`

Answer 2

`"The Minima is "-1/18", and, the Maxima is "1/18.`

Explanation:

Let `x=9tan theta, theta in RR-{(2n+1)pi/2 : n in ZZ}`

Then, `f(x)=(9tantheta)/(81tan^2theta+81)=(9tantheta)/(81sec^2theta)`

`=1/9.sintheta/costheta*cos^2theta=1/9sinthetacostheta`, or,

`=1/18(2sinthetacostheta)=1/18sin2theta`

But, we know that,

`AA theta in RR-{(2n+1)pi/2 : n in ZZ}, -1 le sin2theta le 1`.

Clearly, `"the Minima is "-1/18", and, the Maxima is "1/18.`