Find the local ma

f (x) = x³ – 6x² + 9x + 15

⇒ f’(x) = 3x² – 12x + 9

Now, f’(x) = 0

⇒ 3x² – 12x + 9 = 0

⇒ 3(x-1)(x-3) = 0

⇒ x = 1,3

g’’(x) = 6x – 12 =6(x-2)

Now, f’(1) = 6(1-2)=-6 < 0

and f’(3) = 6(3-2) = 6 > 0

Then, by second derivative test,

⇒ x = 1 is point of local maxima and local maximum of f at x = 1 is

f(1) = 1³ – 6 +9 +15 = 19

And,

x = 3 is point of local minima and local minimum value of f at x = 3 is

f(3) = 27 – 54 + 27 +15 = 15