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The function $f(x) = x^3 + 3x$ is increasing in interval: |
$(-\infty, 0)$ $(0, \infty)$ $\mathbb{R}$ $(0, 1)$ |
$\mathbb{R}$ |
The correct answer is Option (3) → $\mathbb{R}$ ## $f(x) = x^3 + 3x$ $f'(x) = 3x^2 + 3$ For increasing, $f'(x) > 0$ $3x^2 + 3 > 0 \quad (x \in \mathbb{R} : x^2 > 0)$ |
