Show that the function $f$ given by $f(x) = x^3 - 3x^2 + 4x, x \in \mathbb{R}$ is increasing on $\mathbb{R}$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $f'(x) > 0$ for all $x \in \mathbb{R}$, so $f$ is strictly increasing. ##
Here,
$ f'(x) = 3x^2 - 6x + 4 $
$ = 3(x^2 - 2x + 1) + 1 $
$ = 3(x - 1)^2 + 1 > 0, \text{ in every interval of } \mathbb{R} $
Square of any number is always positive.
Therefore, the function $f$ is increasing on $\mathbb{R}$.