critical numbers?

Question

f(x)=xe^-3x

Answer 1

Use the product rule to find the derivative:

u = x
du = 1
v = e^-3x
dv = -3e^-3x

Then you get:

-3xe^(-3x) + e^(-3x)

Reduce that and set equal to 0:

e^(-3x) (1 + 9x) = 0

1 + 9x = 0
9x = -1
x = (-1 / 9)

Answer 2

First, take the derivative of the function.
f'(x)= -3xe^(-3x) + e^(-3x)

Then, take another derivative.
f"(x)= 9xe^(-3x) + -3e^(-3x) + -3e^(-3x)

Set this equal to 0.

9xe^(-3x) + -3e^(-3x) + -3e^(-3x) = 0
x=2/3

Plug this number into a sign chart:

<-----|----->
2/3

Choose a number greater than 2/3 and less than 2/3, then write down whether or not the number is positive or negative.
- +
<-----|----->
2/3

Because the concavity changes, 2/3 is a critical point (the only critical point).

Answer 3

f(x)=xe^-3x
dy/dx= e^-3x -3xe^-3x = e^-3x(1-3x)
So 3x-1=0 --> x=1/3 is a value where function has a maximum.
x= 0 is a root of the equation.

Answer 4

1) take the derivate
2) Set the derivate to 0
3) solve for x

the way you take the derivate of e to the x is:
e^x * d/dx