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

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- May 18th 2010, 01:00 PMeuclid2increasing or decreasing function
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

- May 18th 2010, 01:15 PMArchie Meade
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.