Check whether f(x) = 1- cube root of (x²) satisfies

Check whether $f(x) = 1-\sqrt[3]{x^{2}}$ satisfies the conditions of Rolle's theorem on $[-1,1]$.

Step 1: Continuity

The cube root is defined for every real number, so $f(x) = 1-x^{2/3}$ is defined and continuous on the whole line, in particular on $[-1,1]$. The first condition holds.

Step 2: Endpoint values

\[ f(-1) = 1-\bigl((-1)^{2}\bigr)^{1/3} = 1-1 = 0, \qquad f(1) = 1-1 = 0. \]

So $f(-1) = f(1)$ and the third condition holds too.

Step 3: Differentiability — where it breaks

Away from the origin, $f'(x) = -\frac23 x^{-1/3} = -\frac{2}{3\sqrt[3]{x}}$.

At $x = 0$ the difference quotient is

\[ \frac{f(h)-f(0)}{h} = \frac{-h^{2/3}}{h} = -h^{-1/3} = -\frac{1}{\sqrt[3]{h}}, \]

\[ h\to0^{+}:\ -\frac{1}{\sqrt[3]{h}} \to -\infty; \qquad h\to0^{-}:\ -\frac{1}{\sqrt[3]{h}} \to +\infty. \]

The limit does not exist, so $f$ is not differentiable at $x = 0$, an interior point of $(-1,1)$. The second condition fails.

Step 4: The conclusion fails as well

This is not a case where the theorem's conclusion survives despite a broken hypothesis. For $x\neq0$, $f'(x) = -\frac{2}{3\sqrt[3]{x}} \neq 0 \ \text{ for every } x$.

since a fraction with non-zero numerator never vanishes; and at $x = 0$ the derivative does not exist. Hence there is no $\xi\in(-1,1)$ with $f'(\xi) = 0$.

Step 5: Verification

\[f\bigl(\pm1\bigr) = 0,\quad f(0) = 1 \ \text{(the maximum)},\quad f\bigl(\pm0.001\bigr) = 1-0.01 = 0.99\]

\[f'\bigl(0.001\bigr) = -\frac{2}{3\cdot0.1} \approx -6.67, \qquad f'\bigl(-0.001\bigr) \approx +6.67 \;\;\checkmark\]

\[\text{The slopes blow up in opposite directions: the graph has a cusp at the origin} \;\;\checkmark\]