Question

What are all relative extrema to
f (x)=×^4-32x+4?

Answer 1

`f(x)=x^4-32x+4\\ f'(x)=4x^3-32\\\\ 4x^3-32=0\\ 4(x^3-8)=0\\ x=2\\\\ f(2)=2^4-32\cdot2+4=16-64+4=-44`

The only extremum is
`-44` and it's minimum.