Question

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

Answer 1

`"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.!