Question

Using the second derivative test, how do you find the local maximum and local minimum for `f(x) = x^(3) - 6x^(2) + 5`?

Answer 1

Maxima at x=0, Minima at x=4

Start finding the critical points by equating f '(x)=0

f '(x)= `3x^2` -12x

Critical points would be known by solving `3x^2` -12x=0. This gives x=0, x=4

To apply the second derivative test, find the second derivative f"(x) = 6x-12.

For x=0, f"(x) = -12 (<0), hence it would be a maxima and for x=4, f"(x)=24-12= 12 (>0), it would be a minima.