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_1color(white)(aaaaa)I_2color(white)(aaaa)I_3color(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)nncolor(white)(aaaa)uucolor(white)(aaaa)nncolor(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]}