# Solution ## 1. Find the Critical Points To find the critical points, we first need to find the derivative of the function \( f(x) = x^ + 3x^2 - 5 + 1 \). ### Derivative \[ f'(x) = \frac{}{dx}(x3 + 3x2 - 5x + 1) 3x^2 + 6x - 5 \] Set the derivative to zero to find the critical points: \[ 3x^2 + 6x - 5 = \] ### Solving the Quadratic Equation Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 6 \), and \( c = -5 \): \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \times 3 \times (-5)}}{2 \times 3} \] \[ x = \frac{-6 \pm \sqrt{36 + 60}}{6} \] \[ x = \frac{-6 \pm \sqrt{96}}{6} \] \[ x = \frac{-6 \pm 4\sqrt{6}}{6} \] \[ x = -1 \pm \frac{2\sqrt{6}}{3} \] The critical points are: \[ x_1 = -1 + \frac{2\sqrt{6}}{3} \] \[ x_2 = -1 - \frac{2\sqrt{6}}{3} \] ## 2. Determine the Nature of Each Critical Point We use the second derivative test to determine the nature of each critical point. ### Second Derivative \[ f''(x) = \frac{d}{dx}(3x^2 + 6x - 5) = 6x + 6 \] ### Evaluate the Second Derivative at Each Critical Point For \( x_1 = -1 + \frac{2\sqrt{6}}{3} \): \[ f''(x_1) = 6\left(-1 + \frac{2\sqrt{6}}{3}\right) + 6 \] If \( f''(x_1) > \), then \( x_1 \) is a local minimum. For \( x_2 = -1 - \frac{2\sqrt{6}}{3} \): \[ f''(x_2) = 6\left(-1 - \frac2\sqrt{6}}{3}\right) + 6 \] If \( f''(x_2) < \), then \( x_2 \) is a local maximum. ## 3. Analyze the Behavior as \( x \to \infty \) and \( x \to -\infty \) The leading term \( x^3 \) dominates the behavior of \( f(x) \) as \( x \to \infty \) and \( x \to -\infty \). ### As \( x \to \infty \) \[ f(x) \to \infty \] ### As \( x \to -\infty \) \[ f(x) \to -\infty \] Thus, the function rises to infinity as \( x \) approaches positive infinity and falls to negative infinity as \( x \) approaches negative infinity.