Show that f'(x) = 0 has three real roots for f(x) = x

Show that $f'(x) = 0$ has three real roots for $f(x) = x\bigl(x-1\bigr)\bigl(x-2\bigr)\bigl(x-3\bigr)$, and locate the intervals containing them.

Step 1: The four roots of $f$

\[ f(0) = f(1) = f(2) = f(3) = 0. \]

Step 2: Rolle's theorem on each consecutive pair

$f$ is a polynomial, so on each of $[0,1]$, $[1,2]$, $[2,3]$ it is continuous, differentiable inside, and takes equal (zero) values at the endpoints. Rolle's theorem therefore supplies

\[ \xi_{1}\in\bigl(0,1\bigr), \qquad \xi_{2}\in\bigl(1,2\bigr), \qquad \xi_{3}\in\bigl(2,3\bigr), \]

each with $f'(\xi_{k}) = 0$.

Step 3: There are no others

The three intervals are disjoint, so $\xi_{1},\xi_{2},\xi_{3}$ are distinct. But $f$ has degree $4$, so $f'$ has degree $3$ and can have at most three roots. Hence these are all of them, and each is simple.

Step 4: Finding them explicitly

Pair the factors symmetrically about $x = \tfrac32$. With $u = x^{2}-3x$: \[f = \bigl(x^{2}-3x\bigr)\bigl(x^{2}-3x+2\bigr) = u\bigl(u+2\bigr) = u^{2}+2u,\]

\[ f' = \bigl(2u+2\bigr)u' = 2\bigl(x^{2}-3x+1\bigr)\bigl(2x-3\bigr). \]

Step 5: Verification

\[2x-3 = 0 \;\Longrightarrow\; \xi_{2} = \frac32 = 1.5 \in \bigl(1,2\bigr) \;\;\checkmark\]

\[x^{2}-3x+1 = 0 \;\Longrightarrow\; x = \frac{3\pm\sqrt5}{2}:\ \ \xi_{1} \approx 0.38197,\ \ \xi_{3} \approx 2.61803 \;\;\checkmark\]

\[0\lt 0.382\lt 1\lt 1.5\lt 2\lt 2.618\lt 3 \;\;\checkmark\]