# Solution Given the function: \[ f(x) = x^3 + 3x^2 - 5x + 1 \] ## 1. Find the Critical Points Critical points occur where \( f'(x) = 0 \) or \( f'(x) \) is undefined. First, compute the derivative: \[ f'(x) = 3x^2 + 6x - 5 \] Set the derivative equal to zero: \[ 3x^2 + 6x - 5 = 0 \] Solve this quadratic equation: \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 3 \cdot (-5)}}{2 \cdot 3} \] \[ x = \frac{-6 \pm \sqrt{36 + 60}}{6} \] \[ x = \frac{-6 \pm \sqrt{96}}{6} \] \[ \sqrt{96} = 4\sqrt{6} \] So, \[ x = \frac{-6 \pm 4\sqrt{6}}{6} \] \[ x = \frac{-3 \pm 2\sqrt{6}}{3} \] **Critical points:** \[ x_1 = \frac{-3 + 2\sqrt{6}}{3}, \quad x_2 = \frac{-3 - 2\sqrt{6}}{3} \] --- ## 2. Classify Each Critical Point Use the second derivative test: \[ f''(x) = 6x + 6 \] - At \( x_1 = \frac{-3 + 2\sqrt{6}}{3} \): \[ f''(x_1) = 6\left(\frac{-3 + 2\sqrt{6}}{3}\right) + 6 = -6 + 4\sqrt{6} + 6 = 4\sqrt{6} \] Since \( 4\sqrt{6} > 0 \), **local minimum**. - At \( x_2 = \frac{-3 - 2\sqrt{6}}{3} \): \[ f''(x_2) = 6\left(\frac{-3 - 2\sqrt{6}}{3}\right) + 6 = -6 - 4\sqrt{6} + 6 = -4\sqrt{6} \] Since \( -4\sqrt{6} < 0 \), **local maximum**. --- ## 3. Behavior as \( x \to \infty \) and \( x \to -\infty \) For large \( |x| \), the cubic term \( x^3 \) dominates. - As \( x \to \infty \), \( f(x) \to \infty \) - As \( x \to -\infty \), \( f(x) \to -\infty \) --- ## **Summary Table** | Critical Point | Nature | Second Derivative | |--------------------------------------------------|----------------|-------------------| | \( x_1 = \frac{-3 + 2\sqrt{6}}{3} \) | Local Minimum | \( > 0 \) | | \( x_2 = \frac{-3 - 2\sqrt{6}}{3} \) | Local Maximum | \( < 0 \) | **End Behavior:** - \( x \to \infty, f(x) \to \infty \) - \( x \to -\infty, f(x) \to -\infty \)