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

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Answer 1 · 2015-11-01T11:11:16

$\pm(2sqrt(2))/3$

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

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

Put these three things along and you have

$d/dx 3x-arcsin(x)=3-1/sqrt(1-x^2)$

Now we must find its zeroes:

$3-1/sqrt(1-x^2)=0$

$\iff$

$(3sqrt(1-x^2)-1)/sqrt(1-x^2)=0$

$\iff$

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

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

We can easily solve this last equation:

$3sqrt(1-x^2)=1$

$\iff$

$sqrt(1-x^2)=1/3$

$\iff$

$1-x^2 = 1/9$

$iff$

$x^2 = 8/9$

$iff$

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

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

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