The discussion revolves around the Mean Value Theorem and Rolle's Theorem, specifically analyzing the function f(x) = 1 - x2/3. The original poster seeks to understand why f(-1) equals f(1) while there is no number c in the interval (-1, 1) such that f'(c) equals 0, and how this situation relates to the conditions of Rolle's Theorem.
The discussion is ongoing, with participants exploring the conditions required for Rolle's Theorem to apply. Some participants are questioning the differentiability of the function and whether it meets the necessary criteria for the theorem.
There is a mention of potential nondifferentiable points within the interval, which may affect the application of Rolle's Theorem. Participants are also reflecting on the implications of their calculations and interpretations of the derivative.
For a function to violate Rolle's theorem, it would need to do two things:NanaToru said:Homework Statement
Let f(x) = 1 - x2/3. Show that f(-1) = f(1) but there is no number c in (-1,1) such that f'(c) = 0. Why does this not contradict Rolle's Theorem?
Homework Equations
The Attempt at a Solution
f(x) = 1 - x2/3.
f(-1) = 1 - 1 = 0
f(1) = 1 - 1 = 0
f' = 2/3 x -1/3.
I don't understand why this doesn't have a number c in f'(c), or why Rolle's theorem excludes nondifferentiable points?
If you think ##\frac 1 0 = 0## then you must also think ##0\cdot 0 = 1##?NanaToru said:Is it not 0? Or is that not a valid answer?
Just wait!NanaToru said:...This is honestly the most embarrassing moment of my life. Thank you though!