for f(x)=(x^2)^(1/3)
2006-08-03 04:54:08 · 7 answers · asked by c2pre 2 in Science & Mathematics ➔ Mathematics
This function is not differentiable at x=0.
2006-08-03 05:02:15 · answer #1 · answered by mathematician 7 · 0⤊ 0⤋
The derivative of that function is: f`(x) = 2/(3*x^1/3) which is asymptotic to the x axis. Therefore, f`(x) can never by equal to zero, which means that Rolles' Theorem does not apply to this interval (i.e. the function is non-continuous over this interval)
2006-08-03 12:07:17 · answer #2 · answered by Mr__Roarke 2 · 0⤊ 0⤋
Mathematician and Mr. Roarke are both correct. Judging from their avatars, they are also brothers. Why did they go their separate ways--one into a secret government agency, and the other into the Weather Underground? What will happen when they meet on opposite sides of the law?
2006-08-03 13:05:12 · answer #3 · answered by Benjamin N 4 · 0⤊ 0⤋
No se si me podrias repetir la pregunta mas clara porque no le cacho pero creo que si se la respuesta....es porque es solo una hipotesis lo del intervalo [-1,1]...
2006-08-03 14:21:10 · answer #4 · answered by williamcito 1 · 0⤊ 0⤋
there is no x between -1 and 1 for which the derivative of x is 0.
http://en.wikipedia.org/wiki/Rolle's_theorem
2006-08-03 12:01:25 · answer #5 · answered by Anonymous · 0⤊ 0⤋
Elementary, the hypotenuse is fictitious. Coordinate the duodenum with a proportionate flange.
2006-08-03 11:59:08 · answer #6 · answered by mickeycushman 2 · 0⤊ 0⤋
according to you 0 is not the answer?
2006-08-03 12:07:44 · answer #7 · answered by Caro G 1 · 0⤊ 0⤋