CUET Preparation Today
Show that the function $f$ given by $f(x) = x^3 - 3x^2 + 4x, x \in \mathbb{R}$ is increasing on $\mathbb{R}$. |
$f'(x) > 0$ for all $x \in \mathbb{R}$, so $f$ is strictly increasing. $f'(x) < 0$ for all $x \in \mathbb{R}$, so $f$ is strictly decreasing. $f'(x) = 0$ at $x = 1$, so $f$ is neither increasing nor decreasing. $f'(x) > 0$ only for $x > 1$, so $f$ is increasing for $x > 1$. |
$f'(x) > 0$ for all $x \in \mathbb{R}$, so $f$ is strictly increasing. |
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 $\mathbb{R}$. |
