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

Question

7879 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-01T11:11:16

$\pm \frac{2 \sqrt{2}}{3}$ $\pm(2sqrt(2))/3$

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

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

Put these three things along and you have

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

Now we must find its zeroes:

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

$\Leftrightarrow$ $\iff$

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

$\Leftrightarrow$ $\iff$

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

(provided $x \in (- 1, 1)$ $x\in(-1,1)$ )

We can easily solve this last equation:

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

$\Leftrightarrow$ $\iff$

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

$\Leftrightarrow$ $\iff$

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

$\Leftrightarrow$ $iff$

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

$\Leftrightarrow$ $iff$

$x = \pm \sqrt{\frac{8}{9}} = \pm \frac{2 \sqrt{2}}{3}$ $x=\pm sqrt(8/9) = \pm(2sqrt(2))/3$

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

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