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

Answer 1

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

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

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.
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Answer 2

To find the critical points of ( f(x) = 3x - \arcsin(x) ), we need to find where its derivative is equal to zero or undefined.

( f'(x) = 3 - \frac{1}{\sqrt{1-x^2}} )

Setting ( f'(x) ) equal to zero and solving for ( x ):

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

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

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

( 1-x^2 = \frac{1}{9} )

( x^2 = 1 - \frac{1}{9} )

( x^2 = \frac{8}{9} )

( x = \pm \frac{\sqrt{8}}{3} )

Thus, the critical points of ( f(x) = 3x - \arcsin(x) ) are ( x = \frac{\sqrt{8}}{3} ) and ( x = -\frac{\sqrt{8}}{3} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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