The function $f\left(x\right)=x^3-3x$ is….

• Correct answer
  A

  Increasing in $\left(-\infty ,-1\right)\cup \left(1,\infty \right)$ and decreasing in $\left(-\mathrm{1,1}\right)$
• B

  Increasing in $\left(0,\infty \right)$ and decreasing in $\left(-\infty ,0\right)$
• C

  Decreasing in $\left(0,\infty \right)$ and increasing in $\left(-\infty ,0\right)$
• D

  Decreasing in $\left(-\infty ,-1\right)\cup \left(1,\infty \right)$ and increasing in $\left(-\mathrm{1,1}\right)$

Solution

We have, function $f\left(x\right)=x^3-3x$
$\Rightarrow f^{\text{'}}\left(x\right)=3x^2-3=3\left(x+1\right)\left(x-1\right)$

Here, $f^{\text{'}}\left(x\right)>0in\left(-\infty ,-1\right)\cup \left(1,\infty \right)$ and $f^{\text{'}}\left(x\right)<0$ in (-1, 1).
$\therefore f\left(x\right)$ is increasing in $\left(-\infty ,-1\right)$ , $\cup \left(1,\infty \right)$ and decreasing in (-1, 1).

Hence, the correct answer is option: 'A'.