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.