Find the turning points of the function \(f(x) = x^3 - 3x^2 - 9x + 5\).
Answer
Step 5: Substitute \(x = -1\) and \(x = 3\) into \(f(x)\) to get the corresponding \(y\) values. We get \(f(-1) = -8\) and \(f(3) = -10\). So, the turning points are (-1, -8) and (3, -10).
Step 1 ：Step 1: Find the derivative of the function, \(f'(x) = 3x^2 - 6x - 9\).
Step 2 ：Step 2: Set the derivative equal to zero and solve for \(x\). Thus, \(3x^2 - 6x - 9 = 0\). Solving this quadratic equation gives us \(x = -1, 3\).
Step 3 ：Step 3: Determine whether these points are local minima, local maxima, or saddle points. Take the second derivative of \(f(x)\), which is \(f''(x) = 6x - 6\).
Step 4 ：Step 4: Substitute \(x = -1\) and \(x = 3\) into \(f''(x)\). We get \(f''(-1) = -12\) and \(f''(3) = 12\). Since \(f''(-1) < 0\), \(x = -1\) is a local maximum. Since \(f''(3) > 0\), \(x = 3\) is a local minimum.
Step 5 ：Step 5: Substitute \(x = -1\) and \(x = 3\) into \(f(x)\) to get the corresponding \(y\) values. We get \(f(-1) = -8\) and \(f(3) = -10\). So, the turning points are (-1, -8) and (3, -10).
