How do you find the critical points for #y=x-sqrtx#?

Answer 1

Christopher Agresta

A critical number is a number in the domain at which the derivative is #0# or fails to exist.

For #f(x) = x-sqrtx#, the domain is #[0,oo)#

#f'(x) = 1-1/(2sqrtx) = (2sqrtx-1)/(2sqrtx)#

#f'(x) = 0# #" "# #" " # #" "# #f'(x)# does not exist

#2sqrtx-1=0# #" "# #" "# #" ""# #2sqrtx=0#

#sqrtx = 1/2# #" "# #" "# #" "# #" "# #x=0#

#x = 1/4#

Both #0# and #1/4# are in the domain so both are critical numbers.

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Answer 2

Christopher Agresta

To find the critical points of ( y = x - \sqrt{x} ), you first need to find its derivative, then solve for where the derivative equals zero.

Find the derivative of ( y = x - \sqrt{x} ): [ y' = 1 - \frac{1}{2\sqrt{x}} ]
Set the derivative equal to zero and solve for ( x ): [ 1 - \frac{1}{2\sqrt{x}} = 0 ]
Solve for ( x ): [ \frac{1}{2\sqrt{x}} = 1 ] [ \sqrt{x} = \frac{1}{2} ] [ x = \left(\frac{1}{2}\right)^2 = \frac{1}{4} ]

So, the critical point is ( x = \frac{1}{4} ).

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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.