Question
Show that the function f given by f(x) = x3 – 3x2 + 4x , x ∊ R is strictly increasing on R.
Solution
Here f (x) = x3 – 3x2 + 4x
Df = R
f '(x) = 3x2 – 6x + 4 = 3 (x2 – 2x + 1) + 1 = 3 (x – 1)2 + 1 > 0 ∀ x ∊ R
∴ f is strictly increasing on R.
