Mean Value Theorem/Rolle's Theorem and differentiability

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SUMMARY

The discussion centers on the function f(x) = 1 - x2/3 and its implications regarding Rolle's Theorem. It is established that f(-1) = f(1) = 0, yet there is no value c in the interval (-1, 1) such that f'(c) = 0 due to the non-differentiability of f at x = 0. This situation does not contradict Rolle's Theorem because the function fails to meet the theorem's requirements of differentiability on the open interval.

PREREQUISITES
  • Understanding of Rolle's Theorem
  • Knowledge of differentiability and non-differentiable points
  • Familiarity with basic calculus concepts, including derivatives
  • Ability to analyze piecewise functions and their behavior
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  • Study the conditions and requirements of Rolle's Theorem in detail
  • Learn about differentiability and points of non-differentiability in functions
  • Explore examples of functions that satisfy or violate Rolle's Theorem
  • Investigate the implications of non-differentiable points on calculus theorems
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Students and educators in calculus, particularly those studying the implications of the Mean Value Theorem and Rolle's Theorem, as well as anyone seeking to deepen their understanding of differentiability in mathematical functions.

NanaToru
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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?
 
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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?
For a function to violate Rolle's theorem, it would need to do two things:
1. Satisfy all the hypotheses of Rolle's theorem.
2. Fail to satisfy the conclusion of Rolle's theorem.

Does this function do those two things?
 
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Hm... I'm not sure it satisfies all the hypotheses--from what the back of the book says, it isn't differentiate on the interval of (-1, 1) but I'm not sure how? Did I do a bad job differentiating it?
 
What do you get for f'(0)?
 
Is it not 0? Or is that not a valid answer?
 
NanaToru said:
Is it not 0? Or is that not a valid answer?
If you think ##\frac 1 0 = 0## then you must also think ##0\cdot 0 = 1##?
 
...This is honestly the most embarrassing moment of my life. Thank you though!
 
NanaToru said:
...This is honestly the most embarrassing moment of my life. Thank you though!
Just wait!
 
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