We have,
f(x) = x3 – 6x2 + 9x + 15
Differentiate w.r.t x, we get,
f ‘(x) = 3x2 – 12x + 9
= 3(x2 – 4x + 3)
= 3(x – 3)(x – 1)
For all maxima and minima,
f ’(x) = 0 = 3(x – 3)(x – 1) = 0
= x = 3, 1
At x = 1
f ’(x) changes from –ve to + ve
Since,
x = – 1 is a point of Maxima
At x = 3
f ‘(x) changes from –ve to + ve
Since,
x = 3 is point of Minima.
Hence, local max value f(1) = (1)3 – 6(1)2 + 9(1) + 15 = 19
local min value f(3) = (3)3 – 6(3)2 + 9(3) + 15 = 15