Asked by AJ
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.
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.
Answers
Answered by
bobpursley
a) no, one is not allowed to have zero in the denominator.
b. f(x)=1/sqrt(x)
f'= -1/(2 x<sup>-3/2</sup>) yes it exists
c. if an endpoint does not exist, one cannot integrate it.
b. f(x)=1/sqrt(x)
f'= -1/(2 x<sup>-3/2</sup>) yes it exists
c. if an endpoint does not exist, one cannot integrate it.
Answered by
AJ
Thank you so much! but are you sure for part c.?
Answered by
PLZ spare me
so idk the answer
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