Mean Value Theorem Solution 3

SOLUTION 3: We are given the function $ f(x) = x^2-x^{2/3} $ and the interval $ [-1, 8] $. This function is continuous on the closed interval $ [-1, 8] $ since it is the sum of continuous functions $y=x^2$ (polynomial) and $y=x^{2/3}=(x^2)^{1/3}$ (the functional composition of continuous functions $y=x^2$ and $y=x^{1/3}$). The derivative of $f$ is $$ f'(x) = 2x - (2/3) x^{-1/3} = 2x - \displaystyle{ 2 \over 3 x^{1/3} } $$ We can now see that $f$ is NOT differentiable on the open interval $ (-1, 8) $ since $f'$ is not defined at $x=0$. The assumptions of the Mean Value Theorem have NOT been met, so the Mean Value Theorem does not apply.

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