document.write( "Question 890063: The question is : Discuss applicability of Roll's Theorem for the function
\n" ); document.write( "f(x) = (x)^(2/3) on the interval [-1,1]\r
\n" ); document.write( "\n" ); document.write( "Please send step by stem solution,\r
\n" ); document.write( "\n" ); document.write( "Regards,\r
\n" ); document.write( "\n" ); document.write( "S B Roy \n" ); document.write( "

Algebra.Com's Answer #538764 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
Rolle's theorem CANNOT be applied on the interval [-1,1], because the function is not differentiable at x = 0. The derivative function is $\"df%28x%29%2Fdx+=+2%2F%283x%5E%281%2F3%29%29\"$ which is undefined at x = 0. Thus differentiability on (-1,1) is violated. \r
\n" ); document.write( "\n" ); document.write( "Note that the function $\"f%28x%29+=+x%5E%282%2F3%29\"$ is continuous on the interval [-1,1]. (since the squared cube-root function is continuous), and f(-1) = f(1) = 1. \n" ); document.write( "

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