1. **Problem Statement:** We are given the function $$f(x) = \frac{1}{3}x^3 + (k+1)x^2 + (k^2+5)x$$ and asked to analyze it for different values of the parameter $$k$$ based on the given options: A) $$k > 2$$ B) $$k \leq 2$$ C) $$k < -1$$ D) $$-1 < k < 2$$ 2. **Goal:** We want to understand how the parameter $$k$$ affects the shape or behavior of the cubic function $$f(x)$$. 3. **Key Formula:** The function is a cubic polynomial of the form: $$ f(x) = \frac{1}{3}x^3 + (k+1)x^2 + (k^2+5)x $$ 4. **Step: Find the critical points** To analyze the shape, find the derivative $$f'(x)$$ and solve for critical points where $$f'(x) = 0$$: $$ f'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 + (k+1)x^2 + (k^2+5)x \right) = x^2 + 2(k+1)x + (k^2+5) $$ 5. **Step: Solve the quadratic equation for critical points:** $$ x^2 + 2(k+1)x + (k^2+5) = 0 $$ Use the quadratic formula: $$ x = \frac{-2(k+1) \pm \sqrt{[2(k+1)]^2 - 4 \cdot 1 \cdot (k^2+5)}}{2} $$ Simplify inside the square root: $$ \Delta = 4(k+1)^2 - 4(k^2+5) = 4[(k+1)^2 - (k^2+5)] $$ Expand: $$ (k+1)^2 = k^2 + 2k + 1 $$ So: $$ \Delta = 4[k^2 + 2k + 1 - k^2 - 5] = 4(2k - 4) = 8k - 16 $$ 6. **Step: Analyze the discriminant $$\Delta$$:** - If $$\Delta > 0$$, there are two distinct critical points (local max and min). - If $$\Delta = 0$$, there is one critical point (inflection point). - If $$\Delta < 0$$, no real critical points (monotonic function). Set $$\Delta > 0$$: $$ 8k - 16 > 0 \implies 8k > 16 \implies k > 2 $$ Set $$\Delta = 0$$: $$ 8k - 16 = 0 \implies k = 2 $$ Set $$\Delta < 0$$: $$ 8k - 16 < 0 \implies k < 2 $$ 7. **Interpretation:** - For $$k > 2$$, the function has two critical points, so it has a local maximum and minimum, meaning the graph has a typical cubic "S" shape. - For $$k = 2$$, the function has one critical point, indicating a point of inflection. - For $$k < 2$$, no real critical points, so the function is strictly increasing or decreasing (monotonic). 8. **Check the options:** - A) $$k > 2$$: two critical points, typical cubic shape. - B) $$k \leq 2$$: includes one or no critical points. - C) $$k < -1$$: subset of $$k < 2$$, monotonic. - D) $$-1 < k < 2$$: also subset of $$k < 2$$, monotonic. 9. **Final conclusion:** The key boundary is $$k = 2$$ where the nature of critical points changes. **Answer:** The function has two critical points (local max and min) if and only if $$k > 2$$ (Option A).