Question 1032465
{{{f(x)= x^3-3x^2-9x+5}}} ==> f'(x) = {{{3x^2-6x-9 = 3(x^2-2x-3)}}}.

The stationary points are where f'(x) is zero.

Hence let {{{x^2-2x-3 = (x-3)(x+1) = 0}}}.

==> x = 3, -1.

Now f"(x) = 6x-6 ==>  f"(3) = 12 > 0 ==> local min at x = 3.

f"(-1) = -12 < 0 ==> local max at x = -1.

{{{graph( 500, 500, -15, 15, -25, 20, x^3-3x^2-9x+5)}}}