Show that f(x) = x^4-2x² satisfies the conditions

Show that $f(x) = x^{4}-2x^{2}$ satisfies the conditions of Rolle's theorem on $\bigl[0,\sqrt2\,\bigr]$ and find the corresponding $\xi$. Do the same on $\bigl[-\sqrt2,\sqrt2\,\bigr]$.

Step 1: The three hypotheses

Rolle's theorem asks for continuity on the closed interval, differentiability on the open one, and equal values at the endpoints. The first two are automatic here: $f$ is a polynomial, hence continuous and differentiable on the whole line.

Step 2: Endpoint values on $\bigl[0,\sqrt2\,\bigr]$

\[ f(0) = 0, \qquad f\bigl(\sqrt2\bigr) = \bigl(\sqrt2\bigr)^{4}-2\bigl(\sqrt2\bigr)^{2} = 4-4 = 0. \]

So $f(0) = f\bigl(\sqrt2\bigr) = 0$ and all three conditions hold.

Step 3: The derivative and its roots

\[ f'(x) = 4x^{3}-4x = 4x\bigl(x^{2}-1\bigr) = 0 \;\Longrightarrow\; x = -1,\ 0,\ 1. \]

Of these only $x = 1$ lies in the open interval $\bigl(0,\sqrt2\bigr)$: the value $0$ is an endpoint, and $-1$ is outside.

\[\text{On } \bigl[0,\sqrt2\,\bigr]: \quad \xi = 1\]

Step 4: The symmetric interval $\bigl[-\sqrt2,\sqrt2\,\bigr]$

$f$ is even, so $f\bigl(-\sqrt2\bigr) = f\bigl(\sqrt2\bigr) = 0$ and the hypotheses hold again. Now the open interval is $\bigl(-1.414\ldots,1.414\ldots\bigr)$, which contains all three roots of $f'$:

\[\text{On } \bigl[-\sqrt2,\sqrt2\,\bigr]: \quad \xi \in \{-1,\ 0,\ 1\}\]

Step 5: Verification

\[f'(-1) = -4\bigl(1-1\bigr) = 0, \quad f'(0) = 0, \quad f'(1) = 4\bigl(1-1\bigr) = 0 \;\;\checkmark\]

\[f(-1) = f(1) = 1-2 = -1 \ \text{(the two minima)}, \qquad f(0) = 0 \ \text{(the local maximum)} \;\;\checkmark\]

\[\text{Rolle guarantees at least one } \xi; \text{ here the second interval happens to produce three} \;\;\checkmark\]