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The function $f(x)=x^3-3 x$, is |
increasing on $(-\infty,-1] \cup[1, \infty)$ and decreasing on $(-1,1)$ decreasing on $(-\infty,-1] \cup[1, \infty)$ and increasing on $(-1,1)$ increasing on $(0, \infty)$ and decreasing on $(-\infty, 0)$ decreasing on $(0, \infty)$ and increasing on $(-\infty, 0)$ |
increasing on $(-\infty,-1] \cup[1, \infty)$ and decreasing on $(-1,1)$ |
We have, $f(x)=x^3-3 x \Rightarrow f'(x)=3 x^2-3$ For f(x) to be increasing, we must have $f'(x) \geq 0 \Rightarrow 3 x^2-3 \geq 0 \Rightarrow x^2-1 \geq 0 \Rightarrow x \leq-1$ or $x \geq 1$ Hence, f(x) is increasing on $(-\infty,-1] \cup[1, \infty)$ and decreasing on (-1, 1). |
