What are the local extrema of #f(x)= xe^-x#?

1 Answer
Oct 19, 2016

Answer:

#(1,e^-1)#

Explanation:

We need to use the product rule: #d/dx(uv) = u(dv)/dx+v(du)/dx #

# :. f'(x) = xd/dx(e^-x) + e^-x d/dx(x) #
# :. f'(x) = x(-e^-x) + e^-x (1) #
# :. f'(x) = e^-x-xe^-x #

At a min/max #f'(x)=0#
# f'(x)=0 => e^-x(1-x) = 0 #
Now, # e^x > 0 AA x in RR #
# :. f'(x)=0 => (1-x) = 0 => x=1#

# x=1 => f(1)=1e^-1 = e^-1 #

Hence, there a single turning point at #(1,e^-1)#

graph{xe^-x [-10, 10, -5, 5]}