Home > Calculus > Identifying Stationary Points (Critical Points) for a Function

What are the critical values, if any, of #f(x)= x^(3/4) - 2x^(1/4)#?

Answer 1

Cameron Alexander

By the definition I am accustomed to, they are #0# and #4/9#.

A critical value of #f# is a value in the domain of #f# at which #f'# does not exists or #f'(x)# is #0#.

The domain of the function #f(x)=x^(3/4)-2x^(1/4)# is #[0,oo)#.

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

#f'# fails to exist at #x=0# and #f'(x) = 0# at #x=4/9#.

Both #0# and #4/9# are in the domain, so both are critical values.

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Answer 2

Charles Anctil

To find the critical values of ( f(x) = x^{3/4} - 2x^{1/4} ), we first need to find the derivative of the function, set it equal to zero, and solve for ( x ).

( f'(x) = \frac{3}{4}x^{-1/4} - \frac{1}{2}x^{-3/4} )

Setting the derivative equal to zero:

( \frac{3}{4}x^{-1/4} - \frac{1}{2}x^{-3/4} = 0 )

Solving for ( x ):

( \frac{3}{4}x^{-1/4} = \frac{1}{2}x^{-3/4} )

( \frac{3}{2} = 2x^{-3/4} )

( \frac{3}{4} = x^{-3/4} )

( x^{-3/4} = \frac{3}{4} )

( x = \left(\frac{3}{4}\right)^{-4/3} )

( x = \left(\frac{4}{3}\right)^{4/3} )

So, the critical value is ( x = \left(\frac{4}{3}\right)^{4/3} ).

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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.