Find local max and min

Question

 f(x)=(x^3)(e^5x+2)
 f(x)= ln(2x+4)+x^2

Answer 1

Here are the answers:

(1) The first equation does not have local max and local min because the equation will graph exactly like y = x³ that you should have learned from HS. Note: The (e^5x+2) is irrelevant here, as it will act only like a coefficient for all values of x.

(2) Use the first derivative test for the second one. The first derivative of the equation is f'(x) = (4x²+3x+2)/(2x+4), with the critical points at x = -1 - 0.5 sqrt 2 = - 1.7071 (local max) and x = -1 + 0.5 sqrt 2 = - 0.2929 (local min). Note: The graph will have vertical asymptote at x = 2. Therefore it will not have absolute min.)

Hope these help.