Question
Prove that the function f given by $f(x) = x^3 - 3x^2 + 4x$ is strictly increasing on R.
✓
Answer
$f(x) = x^3 - 3x^2 + 4x$
$f'(x) = 3x^2 - 6x + 4$
$= 3(x^2 - 2x) + 4$
$= 3(x^2 - 2x + 1) - 3 + 4$
$=2(\text{x}-1)^2+1>0,\forall\ \text{x}\in\text{R}$
Hence, $f(x)$ is strictly increasing on R.
$f'(x) = 3x^2 - 6x + 4$
$= 3(x^2 - 2x) + 4$
$= 3(x^2 - 2x + 1) - 3 + 4$
$=2(\text{x}-1)^2+1>0,\forall\ \text{x}\in\text{R}$
Hence, $f(x)$ is strictly increasing on R.
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