Question

Explain why Rolle's Theorem does not apply to the function even though there exist \(a\) and \(b\) such that \(f(a)=f(b)\). $$ \begin{array}{l} f(x)=\sqrt{\left(2-x^{2 / 3}\right)^{3}} \\ {[-1,1]} \end{array} $$

Short answer

Although the function \(f(x)=\sqrt{(2-x^{2 / 3})^{3}}\) is continuous on \([-1,1]\) and \(f(-1) = f(1)\), but Rolle's theorem does not apply to this function because it is not differentiable on the entire interval \((-1,1)\), particularly, it's not differentiable at \(x = 0\).

Step-by-step solution

Verify the function is continuous

Begin by verifying the continuity of the function \(f(x)=\sqrt{(2-x^{2 / 3})^{3}}\) on the interval \([-1,1]\). This function is a square root function and hence, it will be continuous wherever its argument \((2-x^{2 / 3})^{3}\) is greater than or equal to 0. When we substitute \(x = -1\) or \(x = 1\) into the function, the argument of the square root is greater than 0 (specifically it equals 1), and so the function is continuous on the interval \([-1,1]\]. This fulfills one requirement of Rolle's Theorem.

Verify \(f(a) = f(b)\)

For \(a = -1\) and \(b = 1\), substitute these into the function to obtain \(f(a) = \sqrt{(2-(-1)^{2 / 3})^{3}} = \sqrt{1} = 1\) and \(f(b) = \sqrt{(2-(1)^{2 / 3})^{3}} = \sqrt{1} = 1\). Therefore, \(f(a) = f(b)\). This satisfies another condition of Rolle's Theorem.

Check the differentiability of function

To compute \(f'(x)\), we implicitly differentiate \(f(x)=\sqrt{(2-x^{2 / 3})^{3}}\) with respect to \(x\) to obtain \(f'(x) = -\frac{3}{2}(2-x^{2 / 3})^{2} x^{-1 / 3}\). However, \(x = 0\) is in the open interval (-1, 1) but makes the denominator of \(f'(x)\) equal to zero, meaning \(f'(x)\) is undefined at \(x = 0\). Therefore, the function is not differentiable over the entire interval (-1, 1). This implies that Rolle's Theorem does not apply to the function as it fails the differentiability requirement.