For what values of x is #f(x)=x^3e^x# concave or convex?

1 Answer
Jul 17, 2018

The function is concave for #x in (-oo,-3-sqrt3)uu(-3+sqrt3,0)# and convex for #x in (-3-sqrt3, -3+sqrt3)uu(0,+oo)#

Explanation:

Calculate the first and second derivative by the product rule

#(uv)'=u'v-uv'#

#f(x)=x^3e^x#

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

#f''(x)=(3x^2+6x)e^x+(x^3+3x^2)e^x#

#=x(x^2+6x+6)e^x#

The points of inflections are when, #f''(x)=0#

#x(x^2+6x+6)e^x#, #e^x>0#

#=>#, #x(x^2+6x+6)=0#

#=>#, #{(x=0),(x^2+6x+6=0):}#

#=>#, #x=(-6+-sqrt(36-24))/(2)=-3+-sqrt3#

There are #3# points of inflections

Therefore,

There are #4# intervals to consider are

#I_1=(-oo,-3-sqrt3)# and #I_2=(-3-sqrt3, -3+sqrt3)# and #I_3=(-3+sqrt3,0)# and #I_4=(0,+oo)#

Let's consider a variation chart

#color(white)(aaaa)##"Interval"##color(white)(aaaaaa)##I_1##color(white)(aaaaa)##I_2##color(white)(aaaa)##I_3##color(white)(aaaa)##I_4#

#color(white)(aaaa)##"sign f''(x)"##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##" f(x)"##color(white)(aaaaaaaa)##nn##color(white)(aaaa)##uu##color(white)(aaaa)##nn##color(white)(aaaa)##uu#

The function is concave for #x in (-oo,-3-sqrt3)uu(-3+sqrt3,0)# and convex for #x in (-3-sqrt3, -3+sqrt3)uu(0,+oo)#

graph{x^3e^x [-10, 10, -5, 5]}