Question

How do you find the relative extrema for `f(x)=(x^4)+(4x^3)-12`?

Answer 1

`f` has global minimum at `x_0=-3` , `f(-3)=-39`

Explanation:

`f(x)=x^4+4x^3-12`, `D_f=RR`

`f'(x)=4x^3+12x^2=4x^2(x+3)`

`f'(x)=0 <=>` `(x=0` , `x=-3)`

`f'(x)>0 <=> 4x^2(x+3)>0` `<=>` `x` `in` `(-3,0)uu(0,+oo)`

and `f` is continuous at `x_0=0` so `f` is strictly increasing in `[-3,+oo)`

`f'(x)<0 <=> 4x^2(x+3)<0` `<=>` `x` `in` `(-oo,-3)`

so `f` is strictly decreasing in `(-oo,-3]`

• `x<-3` `=>` `f(x)>f(-3)=-39` - (`f` decreasing)
• `x>=-3` `=>` `f(x)>=f(-3)=-39` - (`f` increasing)

As a result `f` has relative minimum at `x_0=-3` , `f(-3)=-39` which also is an global minimum.
(you can mention that `f(0)` is a saddle point.)