Problem 25 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine the open intervals on which the function is increasing, decreasing, or constant. $$f(x)=x^{3}-3 x^{2}+2$$

Step-by-Step Solution

Answer

The function \( f(x) = x^{3}- 3x^{2} + 2 \) is decreasing on the interval \(-\infty, 2\) and increasing on the interval \(2, \infty\).

Step 1: Find the Derivative

The first step is to find the derivative of the function \( f(x) = x^{3}- 3x^{2} + 2 \). Using the power rule, the derivative of the function \( f'(x) = 3x^{2}- 6x\).

Step 2: Find the Critical Numbers

The next step is to find the critical numbers of the function by setting the derivative equal to zero and solving for \(x\). \n\nSo, \(0 = 3x^{2}- 6x\). Simplifying the equation yields \(0 = x(3x- 6)\), which means \(x = 0\) or \(x = 2\). These are the critical numbers.

Step 3: Test the Intervals

The critical numbers divide the number line into three intervals, \(-\infty, 0\), \(0, 2\), and \(2, \infty\). To determine if the function increases or decreases over these intervals, we pick a test number in each interval and evaluate the derivative at that number. If the derivative is negative, the function is decreasing. If the derivative is positive, the function is increasing.\n\n- For the interval \(-\infty, 0\), choose \(-1\) as a test number. Plugging \(-1\) into \(f'(x)\) yields a negative number, which means the function is decreasing on this interval. \n\n- For the interval \(0, 2\), choose \(1\) as a test number. Plugging \(1\) into \(f'(x)\) yields a negative number, so the function keeps decreasing on this interval. \n\n- For the interval \(2, \infty\), choose \(3\) as a test number. Plugging \(3\) into \(f'(x)\) yields a positive number, which means the function is increasing on this interval.