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

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Answer 1 · 2016-12-19T17:14:45

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.