For what values of x is f(x)=(2x^2−4x)e^x concave or convex?

1 Answer
Jan 22, 2018

The function is convex for x in (-oo,-1-sqrt3) uu(-1+sqrt3, +oo) and concave for x in (-1-sqrt3, -1+sqrt3)

Explanation:

"Reminder"

(uv)'=u'v+uv'

Calculate the first and second derivatives

f(x)=(2x^2-4x)e^x=2(x^2-2x)e^x

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

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

The inflection points are when f''(x)=0

x^2+2x-2=0

Solving this quadratic equation for x

x=(-2+-sqrt(2^2-4*1*(-2)))/(2)=(-2+-sqrt(12))/(2)

=(-2+-2sqrt(3))/2

=-1+-sqrt3

The roots are

x_1=-1-sqrt3

x_2=-1+sqrt3

We can build the variation chart

color(white)(aaa)color(white)(aaa)"Interval"color(white)(aaa)(-oo, x_1)color(white)(aaa)(x_1, x_2)color(white)(aaa)(x_2, +oo)

color(white)(aaa)color(white)(aaa)"Sign f''(x)"color(white)(aaaaa)+color(white)(aaaaaaaa)-color(white)(aaaaaaa)+

color(white)(aaa)color(white)(aaa)" f(x)"color(white)(aaaaaaaaa)uucolor(white)(aaaaaaaa)nncolor(white)(aaaaaaa)uu

graph{(2x^2-4x)e^x [-17.35, 14.68, -7.95, 8.07]}