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

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Answer 1 · 2016-09-20T19:57:08

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

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

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 · 2016-09-20T20:08:01

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

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.#