for f(x)=x^4-32x+4 find where f'(x)=0, the intervals for which the function increase and decrease, and all the local extrema
Hi euclid2,
$\displaystyle f'(x)=4x^3-32$
$\displaystyle f'(x)=0\ \Rightarrow\ x^3=\frac{32}{4}=8\ \Rightarrow\ x=2$
$\displaystyle f''(x)=12x^2$
This is positive when x=2, hence it is a minimum
and since there is only one solution for the derivative f'(x)=0, it is the only turning point and therefore the absolute minimum,
there are no local maxima.
f''(x)=0 at x=0, but is always positive otherwise, so the graph is concave up.