Find the minimum and maximum values of f(x) =

Find the minimum and maximum values of $f(x) = x^{5}-5x^{4}+5x^{3}+1$ on $\bigl[-1,\,2\bigr]$.

Step 1: The derivative

\[ f'(x) = 5x^{4}-20x^{3}+15x^{2} = 5x^{2}\bigl(x^{2}-4x+3\bigr) = 5x^{2}\bigl(x-1\bigr)\bigl(x-3\bigr). \tag{1} \]

Step 2: Critical points inside the interval

The zeros of $f'$ are $x = 0$ (double), $x = 1$ and $x = 3$. Only $x = 0$ and $x = 1$ lie in $\bigl[-1,2\bigr]$; $x = 3$ is outside.

Step 3: The point $x = 0$

On both $\bigl(-1,0\bigr)$ and $\bigl(0,1\bigr)$ the factors satisfy $5x^{2}\gt 0$, $x-1\lt 0$, $x-3\lt 0$, so $f'\gt 0$ on both sides. The squared factor touches zero without changing sign, so $x = 0$ is not an extremum — it is a stationary inflection, with $f(0) = 1$.

Step 4: The point $x = 1$

On $\bigl(0,1\bigr)$: $x-1\lt 0$, $x-3\lt 0$, so $f'\gt 0$. On $\bigl(1,2\bigr)$: $x-1\gt 0$, $x-3\lt 0$, so $f'\lt 0$. The change from $+$ to $-$ gives a local maximum:

\[ f(1) = 1-5+5+1 = 2. \]

Step 5: The endpoints

\[ f(-1) = -1-5-5+1 = -10, \qquad f(2) = 32-80+40+1 = -7. \]

Step 6: Compare

\[ \begin{array}{c|c|c|c} x & -1 & 0 & 1 \\ \hline f(x) & -10 & 1 & 2 \end{array} \qquad f(2) = -7 \]

The greatest of these is $f(1) = 2$, the least $f(-1) = -10$.

Step 7: Verification

\[f \text{ rises from } -10 \text{ at } x = -1 \text{ through } f(0) = 1 \text{ to } f(1) = 2,\ \text{then falls to } f(2) = -7 \;\;\checkmark\]

\[f(1.5) = 0.15625,\ \text{between } f(1) = 2 \text{ and } f(2) = -7 \;\;\checkmark\]

\[f''(x) = 10x\bigl(2x^{2}-6x+3\bigr):\ f''(1) = -10\lt 0 \ (\text{max}),\ f''(0) = 0 \ (\text{inflection}) \;\;\checkmark\]