How do you determine whether the function f(x) = x^2e^x is concave up or concave down and its intervals?

1 Answer
Apr 18, 2018

You have to calculate inflection point(s), which mean find zeros of the second derivative.

Explanation:

f''(x)=0

f(x)=x^2*e^(x)
f'(x)=2*x*e^x+x^2*e^x
f''(x)=2*e^x+2*x*e^x+2*x*e^x+x^2*e^x=
=e^x(2+4*x+x^2)

f''(x)=0 <=> 2*e^x(1+x+x^2)=0
Remember that e^x is always* positive
=> (1+x+x^2)=0
x_(1,2) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x_(1,2) = \frac{-4 \pm \sqrt{4^2 - 4*2*1}}{2*1}=
=\frac{-4 \pm \sqrt{16 - 8}}{2}=
=\frac{-4 \pm \sqrt{8}}{2}=
=\frac{-4 \pm 2\sqrt{2}}{2}=
=-2 \pm \sqrt{2}

x_1=-2 +\sqrt(2)=-0.59
x_2=-2 -\sqrt(2)=-3.41
now you have 3 intervals:
1. (-infty,x_2)
2. (x_2,x_1)
3. (x_1,infty)

ask yourself if f''(x) is positive or negative on each interval:
(choose a number of the interval and calculate the value,
don't choose x_1 or x_2 because there is 0 as we calculated before).
1. (-infty,-3.41) x=4, f''(-4)>0
2. (-3.41,-0.59) x=-1, f''(-1)<0
3. (-0.59,infty) x=0, f''(0)>0

if f''(x)>0 => f(x) is convex
if f''(x)<0 => f(x) is concave

convex interval: (-infty,-3.41)uu(-0.59,infty)
concave interval: (-3.41,-0.59)

*lim_(x->-infty) f''(x)=0 (we are looking for abs(x) < infty)