If the derivative of f is given by f'(x)=ex-3x2, at which of the following values of x does f have a relative maximum value?

Answer 1

set the derivative equal to zero and solve for x

ex - 3x^2 = 0
x(e - 3x) = 0
x = 0 or x = e/3

use the 2nd derivative test to see which produces the maximum

f ''(x) = e - 6x
f "(0) = e
f "(e/3) = e - 2e which is negative

so the value of x = e/3 produces a relative maximum value of the function.