# How do you use the second derivative test to find the local maximum and minimum for f(x)=x^4-4x^3+4x^2+6?

Apr 16, 2015

There is a local maxima at x=1 and a minima at x=0 and at x=2

Start finding the critical points by equation f '(x)=0
f '(x)= 4${x}^{3}$ -12${x}^{2}$ +8x =4x( ${x}^{2}$ -3x +2)

f '(x)=0 gives x=0, 1,2

Now get f "(x)= $12 {x}^{2} - 24 x + 8$

At x=0, f "(x)= 8 (>0), hence it is a minima there
At x=1, f"(x)= -4 (<0), hence it is a maxima there
At x=2, f"(x)= 8 (>0), hence it is a minima there.