Is #f(x) =e^x/x-x^3-3# concave or convex at #x=-1#?

1 Answer
Oct 1, 2016

#Convex#

Explanation:

To check if the function is convex or concave we have to find#f''(x)#

If #color(brown)(f''(x)>0)# then #color(brown)(f(x))# is #color(brown)(convex)#
If #color(brown)(f''(x)<0)# then #color(brown)(f(x))# is #color(brown)(concave)#

first let us find #color(blue)(f'(x))#
#f'(x)=((e^x)/x)'-(x^3)'-(3)'#
#f'(x)=(xe^x-e^x)/x^2-3x^2-0#
#color(blue)(f'(x)=(xe^x-e^x)/x^2-3x^2)#

Now let us find #color(red)(f''(x))#
#f''(x)= ((xe^x-e^x)'x^2-(x^2)'(xe^x-e^x))/(x^2)^2-6x#
#f''(x)=((e^x+xe^x-e^x)x^2-2x(xe^x-e^x))/x^4-6x#
#f''(x)=(x^3e^x-2x^2e^x-2xe^x)/x^4-6x#
Let us simplify the fraction by #x#

#color(red)(f''(x)=(x^2e^x-2xe^x-2e^x)/x^3-6x)#

Now let us compute #color(brown)(f''(-1)#

#f''(-1)=((-1)^2e^(-1)-2(-1)e^(-1)-2e^(-1))/(-1)^3-6(-1)#
#f''(-1)=(e^(-1)+2e^(-1)-2e^(-1))/(-1)+6#

#color(brown)(f''(-1)=-e^(-1)+6)#
#color(brown)(f''(-1)>0#

So,#f''(x) >0# at #x=-1#

Therefore,#f(x)# is covex at #x=-1#

graph{e^x/x - x^3 -3 [-20, 20, -20, 20]}