Let's solve the problem step by step as requested. Given function: \[ f(x) = \frac{x^2}{x-2} \] --- ### **Step 1: Find critical number(s)** **Explanation:** Critical numbers occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined (but \( f(x) \) is defined). #### **Find the derivative using the quotient rule:** \[ f'(x) = \frac{(x-2) \cdot 2x - x^2 \cdot 1}{(x-2)^2} \] \[ = \frac{2x(x-2) - x^2}{(x-2)^2} \] \[ = \frac{2x^2 - 4x - x^2}{(x-2)^2} \] \[ = \frac{x^2 - 4x}{(x-2)^2} \] \[ = \frac{x(x-4)}{(x-2)^2} \] Set \( f'(x) = 0 \): \[ x(x-4) = 0 \] \[ x = 0 \quad \text{or} \quad x = 4 \] Also, check where \( f'(x) \) is undefined: \[ (x-2)^2 = 0 \implies x = 2 \] But at \( x=2 \), \( f(x) \) is also undefined (division by zero), so \( x=2 \) is not a critical number. **Critical numbers:** \( \boxed{0 \text{ and } 4} \) --- ### **Step 2: Find Local Maximum and Local Minimum values of \( f \)** **Explanation:** Use the First Derivative Test: - Plug values around critical points into \( f'(x) \) to determine increasing/decreasing behavior. #### **Check intervals:** - For \( x < 0 \): pick \( x = -1 \) \[ f'(-1) = \frac{(-1)(-5)}{(-3)^2} = \frac{5}{9} > 0 \quad (\text{increasing}) \] - Between \( 0 < x < 2 \): pick \( x = 1 \) \[ f'(1) = \frac{1 \cdot (1-4)}{(1-2)^2} = \frac{1 \cdot (-3)}{1} = -3 < 0 \quad (\text{decreasing}) \] - Between \( 2 < x < 4 \): pick \( x = 3 \) \[ f'(3) = \frac{3 \cdot (3-4)}{(3-2)^2} = \frac{3 \cdot (-1)}{1} = -3 < 0 \quad (\text{decreasing}) \] - For \( x > 4 \): pick \( x = 5 \) \[ f'(5) = \frac{5 \cdot (5-4)}{(5-2)^2} = \frac{5 \cdot 1}{9} > 0 \quad (\text{increasing}) \] #### **First Derivative Test Conclusion:** - At \( x = 0 \): \( f'(x) \) changes from positive to negative \( \rightarrow \) **local maximum** - At \( x = 4 \): \( f'(x) \) changes from negative to positive \( \rightarrow \) **local minimum** #### **Find values:** - \( f(0) = \frac{0^2}{0-2} = 0 \) (**local maximum**) - \( f(4) = \frac{4^2}{4-2} = \frac{16}{2} = 8 \) (**local minimum**) --- ### **Final Answers** **a) Critical numbers:** \(\boxed{0,\, 4}\) **b)** - **Local Maximum:** at \( x = 0,\, f(0) = 0 \) - **Local Minimum:** at \( x = 4,\, f(4) = 8 \) \[ \boxed{ \text{Local Maximum at } (0, 0) \\ \text{Local Minimum at } (4, 8) } \]