Verify that the Mean Value Theorem can be applied to the function $f(x)=x^{\frac{2}{3}}$ on the interval $[0,1]$. Then find the value of $c$ in the interval that satisfies the conclusion of the Mean Value Theorem. Enter the exact answer.
Solution
Step 1 ：The Mean Value Theorem states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at that point is equal to the average rate of change of the function over the interval.
Step 2 ：We need to check if the function \(f(x)=x^{\frac{2}{3}}\) is continuous on the closed interval [0, 1] and differentiable on the open interval (0, 1).
Step 3 ：The function \(f(x)=x^{\frac{2}{3}}\) is continuous on the closed interval [0, 1].
Step 4 ：The derivative of the function is \(f'(x)=\frac{2}{3x^{\frac{1}{3}}}\).
Step 5 ：The derivative at x=0 is undefined, so the function is not differentiable on the open interval (0, 1).
Step 6 ：\(\boxed{\text{The Mean Value Theorem cannot be applied to the function } f(x)=x^{\frac{2}{3}} \text{ on the interval } [0,1].}\)
