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How do you find the critical values for #f(x)=x^(2/3)+x^(-1/3)#?

Answer 1

Aurora Alvarado

Find values in the domain of #f# at which #f'(x)# does not exist or #f'(x)=0# . See the explanation.

#f(x)=x^(2/3)+x^(-1/3)#

#"Dom"(f) = RR - {0}#

#f'(x) = 2/3x^(-1/3) - 1/3x^(-4/3)#

# =1/3 x^(-4/3) (2x-1)#

# = (2x-1)/(3x^(4/3))#

#f'(x)# does not exist at #x=0#, which is not in #"Dom"(f)#.

#f'(x) = 0# at #x=1/2# which is in #"Dom"(f)#.

The only critical number for #f# is #1/2#.

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Answer 2

Charles Anctil

To find the critical values of ( f(x) = x^{2/3} + x^{-1/3} ), you need to first find the derivative of the function and then solve for values of ( x ) where the derivative equals zero or is undefined.

Find the derivative of ( f(x) ): [ f'(x) = \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(x^{-1/3}) ]

Using the power rule for differentiation: [ f'(x) = \frac{2}{3}x^{-1/3} - \frac{1}{3}x^{-4/3} ]

Set the derivative equal to zero and solve for ( x ): [ \frac{2}{3}x^{-1/3} - \frac{1}{3}x^{-4/3} = 0 ]

[ \frac{2}{3}x^{-1/3} = \frac{1}{3}x^{-4/3} ]

[ 2x^{-1/3} = x^{-4/3} ]

[ 2 = x^{-1/3} \times x^{-1/3} ]

[ 2 = x^{-2/3} ]

[ x^{-2/3} = 2 ]

[ x = 2^{3/2} ]

So, the critical value for ( f(x) ) is ( x = 2^{3/2} ).

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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.