For what values of x does 4x^6-8x^3+18 have a relative minimum

Answer 1

f(x)=4x^6-8x^3+18

The polynomial has a leading coefficient of 4, therefore there is at least one minimum.

A 6th degree polynomial has a maximum of 5 extrema, and a maximum of 3 minima.

The conditions for an extremum at x0 is f'(x0)=0 and f"(x0)≠0.

The conditions for minima are:
f'(x0)=0 and f"(x0)>0.

f(x)=4x^6-8x^3+18
f'(x)=24x^5-24x^2=24x^2(x^3-1)=0
=> f'(x)=0 at x=0 & x=1 (x∈R)
f"(x)=120x^4-48x
f"(0)=0 and f"(1)=120-48=72>0
Therefore f(x) has a single minimum at x=1.
The graph looks like this:
http://img13.imageshack.us/img13/9940/1333888613.png