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

Jun 17, 2018

critical value is $x = 0$
critical point is $\left(0 , 1\right)$

#### Explanation:

Given: $f \left(x\right) = {e}^{2 x} - 2 x {e}^{2 x}$

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

Derivative rules needed:

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

Product rule: $\left(u v\right) ' = u v ' + v u '$

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

$f ' \left(x\right) = 2 {e}^{2 x} - 2 x \left(2 {e}^{2 x}\right) + \left({e}^{2 x}\right) \left(- 2\right)$

$f ' \left(x\right) = 2 {e}^{2 x} - 4 x {e}^{2 x} - 2 {e}^{2 x}$

$f ' \left(x\right) = - 4 x {e}^{2 x}$

To find critical values:

$f ' \left(x\right) = - 4 x {e}^{2 x} = 0$

$x = 0$ and ${e}^{2 x} = 0$

${\cancel{\ln e}}^{2 x} = \ln \left(0\right) = \text{undefined}$

critical value is $x = 0$

$f \left(0\right) = {e}^{0} - 2 \left(0\right) {e}^{0} = 1$

critical point is $\left(0 , 1\right)$