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

To find the local extremum of the function \( f(x) = x^{3} + 3x^{2} - 9x + 1 \), we first need to find the derivative of the function: \[ f'(x) = 3x^{2} + 6x - 9. \] Next, we set the derivative equal to zero to find the critical points: \[ 3x^{2} + 6x - 9 = 0. \] Dividing the entire equation by 3 gives: \[ x^{2} + 2x - 3 = 0. \] Now, we can factor this quadratic equation: \[ (x + 3)(x - 1) = 0. \] Thus, the critical points are: \[ x = -3 \quad \text{and} \quad x = 1. \] To determine whether these points are local minima or maxima, we can use the second derivative test. First, we find the second derivative: \[ f''(x) = 6x + 6. \] Now, we evaluate the second derivative at the critical points: 1. For \( x = -3 \): \[ f''(-3) = 6(-3) + 6 = -18 + 6 = -12, \] which is less than zero, indicating that \( x = -3 \) is a local maximum. 2. For \( x = 1 \): \[ f''(1) = 6(1) + 6 = 6 + 6 = 12, \] which is greater than zero, indicating that \( x = 1 \) is a local minimum. In summary, the local extremum locations for the function \( f(x) \) are: - Local maximum at \( x = -3 \) - Local minimum at \( x = 1 \)