Calculus - Curve Sketching

Question

For the functions f(x) = xe^x

1. The minimum value of the function occurs when x = _____.

2. A point of inflection occurs when x = ______.

Answer 1

1. f(x) = xe^x; f'(x) = e^x(x + 1) = 0; x = -1;

f''(x) = e^x(x + 1) + e^x = e^x(x + 2); f''(-1) = e^-1 > 0

So, at x = -1 we have minimum value: min f(x) = f(-1) = - 1/e

2.f''(x) = (x + 2)e^x = 0; x = - 2. We have inflection point (-2, -2/e²) becouse at x = - 2 f''(x) change sign from left to right

Answer 2

f'(x) = e^x + xe^x = e^x (1+x)

e^x > 0 for all x , so we have a max / min when f'(x)=0 ie x = -1

f''(x) = e^x + e^x (1+x)

when x= -1, f''(x) > 0, so we have a minimum at x = -1

(2)

Point of inflection . f''(x) = 0, and the sign of f''(x) switches moving through the point.

f''(x) = 0 when e^x (1 + 1 + x) =0, giving x = -2

when x = -1.9 , 2 + x > 0 so f''(x) > 0

when x = - 2.1 2 + x < 0 so f''(x) < 0

so at x=-2 f''(x) changes sign through the point, so x=-2 is an inflection point