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Question
Thanks alot,
Explain why the Mean Value Theorem does not apply to the function f(x)= x to the 1/3 on the interval [-1,1]?
The mean value theorem states that f(x) is continuous on the closed interval [-1,1], &
differentiable on the open interval (-1,1). We can clearly see that f(x)=x^(1/3) has
singularity at x=0 .
However, The mean value theorem is still valid in a slightly more general setting. One only needs to assume that f(x) continuous on [-1,1], and that for every x in (-1,1) the limit
Lim [f(x+h)-f(x)]/h
h->0
exists as a finite number or equals +∞ or −∞. If finite, that limit equals f'(x).
Alon.
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Alon Mandes
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Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
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1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
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M.A in Mathematics & Bs.c in Electronics.
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