How do you find the local extrema of f(x)=x^3-6x?

1 Answer
Aug 23, 2016

"Local minima"=f(sqrt2)=-4sqrt2.

"Local maxima"=f(-sqrt2)=4sqrt2.

Explanation:

For local maxima or minima of a function f,

(1): f'(x)=0; (2): f''(x)<0, or, f''(x)>0, "resp."

Now, f(x)=x^3-6x

:. f'(x)=0 rArr 3x^2-6=0 rArr x^2=6/3=2 rArr x=+-sqrt2

Now, f'(x)=3x^2-6 rArr f''(x)=6x

rArr f''(sqrt2)=6sqrt2>0 rArr"f has a local minima at x"=sqrt2, "which is" f(sqrt2)=sqrt2(2-6)=-4sqrt2.

Again, f''(-sqrt2)=-6sqrt2 <0 rArr f(-sqrt2)=4sqrt2 "is local maxima."

Enjoy Maths.!