Question

What are the critical values, if any, of `f(x) = e^(2x) - 2xe^(2x)`?

Answer 1

critical value is `x = 0`
critical point is `(0, 1)`

Explanation:

Given: `f(x) = e^(2x) - 2xe^(2x)`

Critical values are found by finding the first derivative and setting it equal to zero `f'(x) = 0`:

Derivative rules needed:

`(e^u)' = u' e^u => u = 2x; " "u' = 2; " " (e^(2x))' = 2e^(2x)`

Product rule: `(uv)' = u v' + v u'`

Let `u = -2x; " "u' = -2; " "v = e^(2x); " "v' = 2e^(2x)`

`f'(x) = 2e^(2x) -2x(2e^(2x)) + (e^(2x))(-2)`

`f'(x) = 2e^(2x) - 4xe^(2x) - 2e^(2x)`

`f'(x) = - 4xe^(2x)`

To find critical values:

`f'(x) = - 4xe^(2x) = 0`

`x = 0` and `e^(2x) = 0`

`cancel(ln e)^(2x) = ln(0) = "undefined"`

critical value is `x = 0`

`f(0) = e^0 - 2(0)e^0 = 1`

critical point is `(0, 1)`