Let \( f(x)=\frac{1}{3} x^{3}-2 x^{2} \). 1. Find critical numbers of the function. Enter your answer here \( \mathrm{x}= \) Use a comma to separate answers as needed. If the function has no critical numbers, enter none. 2. Find the \( x \)-coordinate of a relative maximum point if any. Enter your answer here \( \mathrm{x}= \). If the function has no relative maximum points, enter none. 3. Find the \( x \)-coordinate of a relative minimum point if any. Enter your answer here \( \mathrm{x}= \) none. 4. Find the \( x \)-coordinate of an inflection pc any. Enter your answer here \( \mathrm{x}= \) . If the function has no inflection points, enter none.

To find the critical numbers of the function \( f(x)=\frac{1}{3} x^{3}-2 x^{2} \), we'll start by finding its derivative: 1. The derivative is \( f'(x) = x^2 - 4x \). To find critical numbers, we set the derivative equal to zero: \[ x^2 - 4x = 0 \] Factoring gives \( x(x - 4) = 0 \), so the critical numbers are \( x = 0 \) and \( x = 4 \). So, \( \mathrm{x}= 0, 4 \). To find the \( x \)-coordinate of relative maximum and minimum points, we can test these critical numbers in the original function: 2. By applying the first derivative test: - For \( x = 0 \): \( f' \) changes from negative to positive, indicating a relative minimum. - For \( x = 4 \): \( f' \) changes from positive to negative, indicating a relative maximum. Hence, the \( x \)-coordinate for the relative maximum point is \( \mathrm{x}= 4 \). 3. The \( x \)-coordinate for the relative minimum point, as determined above, is \( \mathrm{x}= 0 \). Next, we will check the second derivative to find inflection points: 4. The second derivative is \( f''(x) = 2x - 4 \). Setting this equal to zero gives: \[ 2x - 4 = 0 \] Solving for \( x \) gives \( x = 2 \). Therefore, the \( x \)-coordinate of an inflection point is \( \mathrm{x}= 2 \). Final answers: 1. \( \mathrm{x}= 0, 4 \) 2. \( \mathrm{x}= 4 \) 3. \( \mathrm{x}= 0 \) 4. \( \mathrm{x}= 2 \)