How do you find the second derivative test to find extrema for f(x) = 2x^2lnx-5x^2?

1 Answer
Sep 27, 2015

See the explanation.

Explanation:

D_f=R^+

f'=4xlnx+2x^2*1/x-10x=4xlnx-8x=4x(lnx-2)

f''=4lnx+4x*1/x-8=4lnx-4=4(lnx-1)

f'=0 <=> 4x=0 vv lnx-2=0

x=0 !in D_f
lnx=2 => x=e^2

f''(e^2)=4(lne^2-1)=4(2-1)=4>0 and hence function has a minimum at x=e^2.

f_min=f(e^2)=2(e^2)^2lne^2-5(e^2)^2=4e^4-5e^4=-e^4