Critical numbers of f(x) = xe^(2x)?

Question

I know that you need to find derivative and find f'(x) = 0 or dne, i just don't know how to solve it with the [e] in the equation. help anyone?

f(x) = xe^(2x)
f'(x) = (2x + 1)e^(2x)

f(x) = 0
(2x + 1)e^(2x) = 0
e^(2x) = 0
...
that is where i'm lost

Answer 1

First, your last line is incorrect; you were correct up to the line:
(2x + 1)e^(2x) = 0

Therefore 2x + 1 = 0, or e^(2x) = 0, since the product of zero with any other finite number is zero.

Now, e raised to any finite number cannot equal zero (and your critical numbers must be finite). Thus, the only critical point exists when 2x + 1 = 0, or x = -1/2.

Answer 2

This is a continuous function so the derivative won't really help you all that much. This is a function used often in probability. It's limit at infinity is infinity, so it's usually bounded.

x = -1/2 is an answer to your f'(x) = 0
e^ (negative infinity) also equals 0. Think about 2^n for help with this. 2^-1 is a half, and 2^-2 is a quarter. If you kept subtracting one from the exponent, your result would keep halving forever, to the theoretical value of 0.

Answer 3

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

so either (2x+1)=0 or e^(2x)=0

so we have x= -1/2 or x is minus infinity

Answer 4

E is a constant. You need to take the logarithim values of these..Hope that helps!