# What are the critical points of f(x) = 3x-arcsin(x)?

Nov 1, 2015

$\setminus \pm \frac{2 \sqrt{2}}{3}$

#### Explanation:

To find critical points, simply derive and find zeroes of the derivative:

1. The derivative of $3 x$ is $3$;
2. The derivative of $\arcsin \left(x\right)$ is $\frac{1}{\sqrt{1 - {x}^{2}}}$;
3. The derivative of a difference of functions is the difference of the derivatives of the functions.

Put these three things along and you have

$\frac{d}{\mathrm{dx}} 3 x - \arcsin \left(x\right) = 3 - \frac{1}{\sqrt{1 - {x}^{2}}}$

Now we must find its zeroes:

$3 - \frac{1}{\sqrt{1 - {x}^{2}}} = 0$

$\setminus \iff$

$\frac{3 \sqrt{1 - {x}^{2}} - 1}{\sqrt{1 - {x}^{2}}} = 0$

$\setminus \iff$

$3 \sqrt{1 - {x}^{2}} - 1 = 0$

(provided $x \setminus \in \left(- 1 , 1\right)$)

We can easily solve this last equation:

$3 \sqrt{1 - {x}^{2}} = 1$

$\setminus \iff$

$\sqrt{1 - {x}^{2}} = \frac{1}{3}$

$\setminus \iff$

$1 - {x}^{2} = \frac{1}{9}$

$\iff$

${x}^{2} = \frac{8}{9}$

$\iff$

$x = \setminus \pm \sqrt{\frac{8}{9}} = \setminus \pm \frac{2 \sqrt{2}}{3}$

And since $\setminus \pm \sqrt{\frac{8}{9}} \setminus \in \left(- 1 , 1\right)$, we can accept the result.