What are the points of inflection of #f(x)=2xe^-x + x^2e^x
#?

Question

What are the points of inflection of #f(x)=2xe^-x + x^2e^x
  #?

Evelyn Agard · Accepted Answer

#f(x)# has one point of inflection, which is at #x approx .161# #f(x)=2xe^(-x)+x^2e^x# To find points of inflection, find where #f''(x)# changes sign. Use the product rule and chain rule: #f'(x)=(2)(e^(-x))+(2x)(-e^(-x))+(2x)(e^x)+(x^2)(e^x)# Rewrite/simplify: #f'(x)=2e^(-x)-2xe^(-x)+2xe^x+x^2e^x# More differentiation... #f''(x)=-2e^(-x)-(2)(e^(-x))-(2x)(-e^(-x))+(2)(e^x)+(2x)(e^x)+(2x)(e^x)+(x^2)(e^x)# More simplification... #f''(x)=-4e^(-x)+2xe^(-x)+2e^x+4xe^x+x^2e^x# Set #f''(x)# equal to zero using graphing calculator: #0=f''(x)# #x approx .16059649#

Ezekiel Alexander · Answer

undefined To find the points of inflection for the function $ f(x) = 2xe^{-x} + x^2e^x $, we first need to find its second derivative, $ f''(x) $, and then solve for $ x $ where $ f''(x) = 0 $.

First derivative: 
$$ f'(x) = 2e^{-x} - 2xe^{-x} + 2xe^x + 2x^2e^x $$

Second derivative: 
$$ f''(x) = -2e^{-x} - 2xe^{-x} - 2xe^x + 2e^x + 2x^2e^x + 2x^2e^x $$

Simplify the second derivative: 
$$ f''(x) = -2e^{-x} - 2xe^{-x} - 2xe^x + 2e^x + 4x^2e^x $$

Now, set $ f''(x) $ equal to zero and solve for $ x $: 
$$ -2e^{-x} - 2xe^{-x} - 2xe^x + 2e^x + 4x^2e^x = 0 $$

There is no simple analytical solution to this equation. You would typically use numerical methods to approximate the solutions. Once you find the $ x $-values where the second derivative equals zero, you can test these values to determine whether they correspond to points of inflection by checking the concavity of the function around those points. If the concavity changes at those points, they are points of inflection.

What are the points of inflection of #f(x)=2xe^-x + x^2e^x #?