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State why Rolle's Theorem does not apply to f(x)= x^(2/3) on the interval [-1,1]

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Hello Lana,

First, what is Rolle's Theorem?

  "If a real-valued function ƒ is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and ƒ(a) = ƒ(b), then there exists a c in the open interval (a, b) such that f'(c) =0."

  One way to look at this question (especially if you are a visual learner, like me) is to tabulate and graph Y(x) and Y'(x).  (Note:  I use Y for the functionnotation f)

Y=X^(2/3)   Y'=(2/3)*X^(-1/3)

Y:  From x=-1 to x=0 Y decreases from 1 to exactly 0. 

     From x=1 to x =0 Y decreases from 1 to exactly 0 also.  Y is a mirror image on either side of x=0 and abruptly changes slope sign at zero.

  Thus x is continuous on [-1,1], Y(-1)=Y(1) but its derivative is discontinuous at 0.  The first requirement of Rolle's Theorem is not satisfied.

Y':  We can write down Y' as Y'=(2/3)*X^(-1/3).  

From x=-1 to x=0 Y' decreases from about -1.4 to negative infinity at 0.

From x=1 to x =0 Y increase from about +1.4 to positive infinity at zero.  Under these conditions, Y' is discontinuous and Y' never equals zero.    This is what is predictied by Rolle's Theorem based on the information we initally determiined about Y.

Caveat:  Tabulating and graphing Y and Y' can provide insight into a function's characteristics and behavior but care must be taken to look in detail at portions of a function where values are rapidly changing.  If your graph is too coarse, details will be missed.  For Example: Plot Y and Y' above between x= -1 to 1 at intervals of 0.1 and look at the graphs' behaviors.  Next plot the functions between x= -0.1 and +0.1 at intervals of 0.01 (Where all the action occurs!).  You will see the second pair of graphs disclose significant detail missed or at least not apparent in the first graphs.

Bruce S.