With the function f(X) = (x)^(1/2), (square root function), what is the instant rate of change when x =0?
2007-10-24 06:01:11 · 7 answers · asked by Ask or Answer 2 in Science & Mathematics ➔ Mathematics
to get f'(x) the f(x) must be continuous at that point
f(0) is zero but it is an end no any point to the left of it on the graph
so it is not possible to get f'(0)
2007-10-24 06:07:23 · answer #1 · answered by mbdwy 5 · 0⤊ 0⤋
is it undefined?
f'(x)= 1/2 x ^ (-1/2)
when x = 0, f'(x) is 1/0
2007-10-24 13:07:07 · answer #2 · answered by norman 7 · 0⤊ 0⤋
it is undefined, because when you take dervitative of f(x), it gives you a square root of in the denominator.
f'(0) = Undefined
2007-10-24 13:16:21 · answer #3 · answered by 1294 4 · 0⤊ 0⤋
Undefined.
2007-10-24 13:04:35 · answer #4 · answered by fcas80 7 · 0⤊ 0⤋
infinity. It aproaches infinity as x approaches zero.
2007-10-24 13:16:40 · answer #5 · answered by Anonymous · 0⤊ 0⤋
cannot be defined
2007-10-24 13:13:00 · answer #6 · answered by Shahbaz 2 · 0⤊ 0⤋
f' = 0.5x^-0.5
Undefined
2007-10-24 13:08:25 · answer #7 · answered by gebobs 6 · 0⤊ 0⤋