Determine the location of each local extremum of the function. \( f(x)=x^{3}+3 x^{2}-9 x+1 \)

To find the local extrema of the function \( f(x) = x^3 + 3x^2 - 9x + 1 \), we start by calculating the first derivative: \[ f'(x) = 3x^2 + 6x - 9. \] Setting the first derivative to zero helps us find critical points: \[ 3x^2 + 6x - 9 = 0. \] Dividing the entire equation by 3, we get: \[ x^2 + 2x - 3 = 0. \] Factoring yields: \[ (x + 3)(x - 1) = 0, \] which gives us the critical points: \[ x = -3 \quad \text{and} \quad x = 1. \] Next, we find the second derivative to determine the nature of these critical points: \[ f''(x) = 6x + 6. \] Evaluating \( f''(x) \) at the critical points: 1. For \( x = -3 \): \[ f''(-3) = 6(-3) + 6 = -18 + 6 = -12 < 0, \] indicating a local maximum. 2. For \( x = 1 \): \[ f''(1) = 6(1) + 6 = 6 + 6 = 12 > 0, \] indicating a local minimum. Thus, the function has a local maximum at \( x = -3 \) and a local minimum at \( x = 1 \). To find the function values at these critical points: - At \( x = -3 \): \[ f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 1 = -27 + 27 + 27 + 1 = 28. \] - At \( x = 1 \): \[ f(1) = (1)^3 + 3(1)^2 - 9(1) + 1 = 1 + 3 - 9 + 1 = -4. \] In conclusion, the local maximum occurs at \( (-3, 28) \) and the local minimum at \( (1, -4) \).