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

Question

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

Elijah Abram · Accepted Answer

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)# #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)))# 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)#

Christopher Allers · Answer

undefined To determine where the function $ f(x) = x - x^2e^{-x} $ is concave or convex, we need to find its second derivative and analyze its sign:

1. Find the first derivative of $ f(x) $:
$$ f'(x) = 1 - (2x - 2xe^{-x} - x^2(-e^{-x})) $$
$$ f'(x) = 1 - (2x - 2xe^{-x} + x^2e^{-x}) $$
$$ f'(x) = 1 - 2x + 2xe^{-x} - x^2e^{-x} $$

2. Find the second derivative of $ f(x) $:
$$ f''(x) = -2 + 2e^{-x} + 2xe^{-x} - 2xe^{-x} - x^2e^{-x} $$
$$ f''(x) = -2 + 2e^{-x} - x^2e^{-x} $$

3. Now, analyze the sign of $ f''(x) $ to determine concavity or convexity:
$$ f''(x) > 0 $$ for convexity,
$$ f''(x) < 0 $$ for concavity.

Given the expression for $ f''(x) = -2 + 2e^{-x} - x^2e^{-x} $, it's not easy to determine the sign directly. You may use numerical methods or further analysis to find specific values of $ x $ where the function is concave or convex.