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

Question

5033 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-17T16:58:48

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

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)#