Question

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

Answer 1

Find the second derivative and check its sign. It's convex if it's positive and concave if it's negative.

Concave for:
`x in(2-sqrt(2),2+sqrt(2))`

Convex for:
`x in(-oo,2-sqrt(2))uu(2+sqrt(2),+oo)`

Explanation:

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

First derivative:

`f'(x)=1-(2xe^-x+x^2*(-e^-x))`

`f'(x)=1-2xe^-x+x^2e^-x`

Take `e^-x` as a common factor to simplify next derivative:

`f'(x)=1+e^-x*(x^2-2x)`

Second derivative:

`f''(x)=0+(-e^-x*(x^2-2x)+e^-x*(2x-2))`

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

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

Now we must study the sign. We can switch the sign for easily solving the quadratic:

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

`Δ=b^2-4*a*c=4^2-4*1*2=8`

To make the quadratic a product:

`x_(1,2)=(-b+-sqrt(Δ))/(2*a)=(4+-sqrt(8))/(2*1)=2+-sqrt(2)`

Therefore:

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

• A value of `x` between these two solutions gives a negative quadratic sign, while any other value of `x` makes it positive.
• Any value of `x` makes `e^-x` positive.
• The negative sign at the start of the function reverses all signs.

Therefore, `f''(x)` is:

Positive, therefore concave for:
`x in(2-sqrt(2),2+sqrt(2))`

Negative, therefore convex for:
`x in(-oo,2-sqrt(2))uu(2+sqrt(2),+oo)`