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

2 Answers
Sep 20, 2016

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."

Sep 20, 2016

"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.