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

Question

Scarlett Allison · Accepted Answer

#(1,e^-1)#

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]}

Isabella Ackerman · Answer

undefined The local extrema of $ f(x) = x e^{-x} $ occur where its derivative equals zero or where the derivative is undefined.

To find the derivative of $ f(x) $, we use the product rule:

$ f'(x) = e^{-x} - xe^{-x} $

To find where the derivative equals zero:

$ e^{-x} - xe^{-x} = 0 $

Solving this equation gives:

$ x = 1 $

Now, to classify whether this point is a local maximum or minimum, we can check the second derivative:

$ f''(x) = -e^{-x} + xe^{-x} $

Plugging $ x = 1 $ into the second derivative:

$ f''(1) = -e^{-1} + e^{-1} = 0 $

Since the second derivative at $ x = 1 $ is zero, the test fails and we need to use other methods to determine the nature of the extremum. Typically, this involves examining the behavior of the function around $ x = 1 $.

By observing the behavior of the function, we see that it changes from increasing to decreasing around $ x = 1 $, indicating that $ x = 1 $ is a local maximum.

Therefore, the local maximum of $ f(x) = x e^{-x} $ is at $ x = 1 $.