Question
Prove that the function f given by f(x) = x3 - 3x2 + 4x is strictly increasing on R.

Answer

f(x) = x3 - 3x2 + 4x
f'(x) = 3x2 - 6x + 4
= 3(x2 - 2x) + 4
= 3(x2 - 2x + 1) - 3 + 4
$=2(\text{x}-1)^2+1>0,\forall\ \text{x}\in\text{R}$
Hence, f(x) is strictly increasing on R.

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