f (x) = 2 x3 – 6 x2 + 6 x + 5
∴  f ' (x) = 6 x2 – 12 x + 6 = 6 (x2 – 2 x + 1)
∴  f ' (x) = 6 ( x – 1)2
Now f ' (x) = 0 ⇒ 6 (x – 1)2 = 0 ⇒ r = 1
When x < 1 slightly, f ' (x) = 6 (1) = + ve
When x > 1 slightly, f ' (x) = 6 (+) = ve
∴ f ' (x) does not change sign as x passes through I
∴  x = 1 is a point of inflexion.
Find local maximum and local minimum values of the function f given by
f (a) = 3x4 + 4x3 – 12x2 + 12