The function \( f \) is defined as \( f(x)=x^{3}+3 x^{2}-9 x-8 \), where \( x \in \mathbb{R} \). (a) Find \( f^{\prime}(x) \). (b) Solve the equation \( f^{\prime}(x)=0 \). (c) Hence, find the turning points of the curve \( y=f(x) \) and state whether each point is a local maximum or a local minimum.

To solve the given problem step by step: (a) To find the derivative \( f'(x) \): \[ f(x) = x^{3} + 3x^{2} - 9x - 8 \] Using the power rule, we differentiate: \[ f'(x) = 3x^{2} + 6x - 9 \] (b) Next, we solve the equation \( f'(x) = 0 \): \[ 3x^{2} + 6x - 9 = 0 \] Dividing every term by 3 gives: \[ x^{2} + 2x - 3 = 0 \] Now we can factor this quadratic equation: \[ (x + 3)(x - 1) = 0 \] Thus, the solutions are: \[ x = -3 \quad \text{and} \quad x = 1 \] (c) To find the turning points, we substitute these \( x \) values back into \( f(x) \): 1. For \( x = -3 \): \[ f(-3) = (-3)^{3} + 3(-3)^{2} - 9(-3) - 8 = -27 + 27 + 27 - 8 = -8 \] Turning point: (-3, -8) 2. For \( x = 1 \): \[ f(1) = (1)^{3} + 3(1)^{2} - 9(1) - 8 = 1 + 3 - 9 - 8 = -13 \] Turning point: (1, -13) Now we check the nature of these turning points using the second derivative: \[ f''(x) = 6x + 6 \] - For \( x = -3 \): \[ f''(-3) = 6(-3) + 6 = -18 + 6 = -12 \quad (\text{local maximum}) \] - For \( x = 1 \): \[ f''(1) = 6(1) + 6 = 6 + 6 = 12 \quad (\text{local minimum}) \] In summary: - Turning point at (-3, -8) is a local maximum. - Turning point at (1, -13) is a local minimum.