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

1 Answer
Aug 23, 2016

Answer:

#"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.!