Please help me solve this calculus questions.

Question

Consider the function f(x) = x² e^-x, then answer the following questions.

a) Determine the intervals of increase and decrease for f(x).

b) Determine the intervals of concavity for f(x)

c) Determine the coordinates of any points of inflection.

Answer 1

Hi Laura,

inflection points and concavity tell us to focus on the second derivative, while intervals of inc/decreasing tell us to focus on the first. Since we need the second, though, I'll take the first two derivatives before I actually answer the questions.

f(x) = x²e^-x

f'(x) = (x²)(e^-x)(-1) + (e^-x)(2x) (product rule with some chain rule)

= e^-x(2x - x²) **factored out e^-x. We'll use this later.

f ' ' (x) = e^-x(x² - 4x + 2). (product rule with some chain rule in there again)

We know that inflection points are where the second derivative equals zero. I factored out e^-x in the second derivative, and we know it will never be zero since it has an asymptote at y=0 (along the x-axis). Solving x² - 4x + 2 - 0 using the quadratic formula yields x= 2 + sqrt(2), 2 - sqrt(2). Plugging these x-values back into the second derivative will give you the coordinates for the inflection points.

Plugging in x-values around those points (say, x=0, x=2, and x=5) into the second derivative gives you positive, negative, and positive answers--those are your intervals of concavity:

concave up: negative infinity to 2-sqrt(2), 2+sqrt(2) to infinity

concave up: 2-sqrt(2) to 2+sqrt(2)

note that the endpoints will all have parenthesis instead of brackets since they are non-inclusive (infinity is not a number, and inflection points are neither concave up or down since they equal zero)

To find intervals of increasing and decreasing, we do a similar process using f'(x)=0.

e^-x(2x - x²) = 0. Again, e^-x cannot ever equal zero, so 2x - x^2 must equal zero. Solving this gives us x=0 and x=2. Plug values around 0 and 2 (ex: -2, 1, 3) into the first derivative shows that it's increasing on (0,2) and decreasing outside of that.

Hope this helps!