What are the points of inflection of f(x)= x^3 - 7x^2 + 2xe^x f(x)=x37x2+2xex?

1 Answer
May 12, 2017

f(x)f(x) has one point of inflection at x approx .3288x.3288

Explanation:

f(x)=x^3-7x^2+2xe^xf(x)=x37x2+2xex

f'(x)=3x^2-14x+ color(blue)(2xe^x)+2e^x

f''(x)=6x-14+ color(blue)(2xe^x+2e^x)+2e^x

f''(x)=6x-14+ 2xe^x+4e^x

To find points of inflection, set f''(x) equal to zero.
0=2[3x-7+xe^x+4e^x]

(using a graphing calculator):
x approx .3288