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

Find the minimum and maximum values of $\displaystyle f(x) = \frac{5-x}{9-x^{2}}$ on $\bigl[0,2\bigr]$.

Step 1: The function is continuous on the interval

The denominator $9-x^{2} = \bigl(3-x\bigr)\bigl(3+x\bigr)$ vanishes only at $x = \pm3$, and neither point lies in $\bigl[0,2\bigr]$. On the interval $9-x^{2}\ge5\gt 0$, so $f$ is continuous there and attains both a least and a greatest value.

Step 2: The derivative

By the quotient rule, \[f'(x) = \frac{\bigl(-1\bigr)\bigl(9-x^{2}\bigr)-\bigl(5-x\bigr)\bigl(-2x\bigr)}{\bigl(9-x^{2}\bigr)^{2}} = \frac{-9+x^{2}+10x-2x^{2}}{\bigl(9-x^{2}\bigr)^{2}} = \frac{-x^{2}+10x-9}{\bigl(9-x^{2}\bigr)^{2}}, \tag{1}\]

and factoring the numerator, $-x^{2}+10x-9 = -\bigl(x-1\bigr)\bigl(x-9\bigr)$: \[f'(x) = \frac{-\bigl(x-1\bigr)\bigl(x-9\bigr)}{\bigl(9-x^{2}\bigr)^{2}}. \tag{2}\]

Step 3: Critical points

The denominator of (2) is positive on $\bigl[0,2\bigr]$, so $f'(x) = 0$ exactly when $x = 1$ or $x = 9$. Only $x = 1$ belongs to the interval.

Step 4: Sign of the derivative

On $\bigl[0,2\bigr]$ the factor $x-9$ is negative throughout, so the sign of $f'$ is the sign of $-\bigl(x-1\bigr)\cdot\bigl(\text{negative}\bigr) = \bigl(x-1\bigr)$: $0\lt x\lt 1:\ f'\lt 0\ \ (\text{decreasing}); \qquad 1\lt x\lt 2:\ f'\gt 0\ \ (\text{increasing})$.

The derivative changes from $-$ to $+$ at $x = 1$, so that point is a local minimum.

Step 5: Compare the three candidates

\[ f(0) = \frac59 \approx 0.5556, \qquad f(1) = \frac{5-1}{9-1} = \frac48 = \frac12, \qquad f(2) = \frac{5-2}{9-4} = \frac35 = 0.6. \]

The smallest of the three is $f(1)$, the largest $f(2)$.

Step 6: Verification

\[f(0.5) = 0.5142857 \gt f(1) = 0.5; \qquad f(1.5) = 0.5185185 \lt f(2) = 0.6 \;\;\checkmark\]

\[f''(1) = \frac{-2\cdot1+10}{\bigl(9-1\bigr)^{2}} = \frac{8}{64} = \frac18 \gt 0,\ \text{confirming a minimum} \;\;\checkmark\]

\[\text{The maximum is at an endpoint, not a critical point: } f'(2) = \frac{7}{25} \gt 0 \;\;\checkmark\]