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

Question

3955 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-27T06:11:49

See the 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#