Is #f(x)=-xe^x+x^3-x^2# concave or convex at #x=-2#?

Question

1513 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-15T12:23:32

Since #f''(-2)<0#, the function is concave

First, you would calculate the first and the second derivatives:

#f'(x)=(-1*e^x+(-x)*e^x)+3x^2-2x#

#=(-e^x-xe^x)+3x^2-2x#

#f''(x)=(-e^x-e^x-xe^x)+6x-2#

#=-2e^x-xe^x+6x-2#

Then you would calculate

#f''(-2)=-2e^(-2)-(-2)e^(-2)+6(-2)-2#

#=cancel(-2e^-2)cancel(+2e^-2)-12-2#

#-14<0#

Since #f''(-2)<0#, the function is concave
graph{-xe^x+x3-x^2 [-10.035, 9.965, -7.32, 2.68]}