How do you find the second derivative test to find extrema for #f(x) = 2x^2lnx-5x^2#? Calculus Graphing with the Second Derivative First Derivative Test vs Second Derivative Test for Local Extrema 1 Answer Sasha P. 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# Answer link Related questions What is the first derivative test to determine local extrema? How does the first derivative test differ from the second derivative test? How does the first derivative test work? What is the first derivative test for critical points? What is the first derivative test for local extreme values? What does the first derivative test tell you? How do you know when to use the first derivative test and the second derivative test? How do you use the second derivative test to find the local extrema for #f(x)=sin(x)#? How do you use the second derivative test to find the local extrema for #f(x)=e^(x^2)#? How do you use the second derivative test how do you find the local maxima and minima of #f(x) =... See all questions in First Derivative Test vs Second Derivative Test for Local Extrema Impact of this question 3375 views around the world You can reuse this answer Creative Commons License