a) no, one is not allowed to have zero in the denominator.
b. f(x)=1/sqrt(x)
f'= -1/(2 x-3/2) yes it exists
c. if an endpoint does not exist, one cannot integrate it.
Given the function f(x) = 1/(sqrt x)
a. Does f(x) exist at x=0?
b. Does f'(x) exist at x=0?
c. Explain why integral (from 0->1) f(x)dx does exist.
THANK YOU.
3 answers
Thank you so much! but are you sure for part c.?
so idk the answer