1. Key Concepts and Formulas

• Finding Local Extrema: To find local minima/maxima of a function $f(x)$, find critical points where $f'(x) = 0$. Use the second derivative test: if $f''(x) > 0$, it's a local minimum; if $f''(x) < 0$, it's a local maximum.
• Solving Rational Inequalities: To solve $\frac{P(x)}{Q(x)} < 0$, find roots of $P(x)$ and $Q(x)$, then analyze the sign of the expression in the intervals defined by these roots. If $P(x)$ is always positive, the sign depends only on $Q(x)$.

2. Step-by-Step Solution

Step 1: Define the function and find its first derivative. The function is $f(x) = 1 + x(\lambda^2 - x^2)$. We need to find its derivative to locate potential minima/maxima. $f(x) = 1 + \lambda^2 x - x^3$ $f'(x) = \frac{d}{dx}(1 + \lambda^2 x - x^3) = \lambda^2 - 3x^2$ Explanation: We expanded the function and then applied the power rule of differentiation to each term.

Step 2: Find the critical points. Critical points occur where $f'(x) = 0$. $\lambda^2 - 3x^2 = 0$ $3x^2 = \lambda^2$ $x^2 = \frac{\lambda^2}{3}$ $x = \pm \frac{\lambda}{\sqrt{3}}$ Explanation: We set the derivative equal to zero and solved for $x$. These are the x-values where a local minimum or maximum could occur.

Step 3: Use the second derivative test to determine the local minimum. We find the second derivative to determine the nature of the critical points. $f''(x) = \frac{d}{dx}(\lambda^2 - 3x^2) = -6x$ Now, we evaluate $f''(x)$ at each critical point:

• At $x = \frac{\lambda}{\sqrt{3}}$: $f''\left(\frac{\lambda}{\sqrt{3}}\right) = -6\left(\frac{\lambda}{\sqrt{3}}\right) = -2\sqrt{3}\lambda$
• At $x = -\frac{\lambda}{\sqrt{3}}$: $f''\left(-\frac{\lambda}{\sqrt{3}}\right) = -6\left(-\frac{\lambda}{\sqrt{3}}\right) = 2\sqrt{3}\lambda$ Since $\lambda > 0$, $f''(\frac{\lambda}{\sqrt{3}}) < 0$, indicating a local maximum. And $f''(-\frac{\lambda}{\sqrt{3}}) > 0$, indicating a local minimum. Therefore, the point of local minimum is at $x = -\frac{\lambda}{\sqrt{3}}$. Explanation: The second derivative test allows us to classify critical points. A positive second derivative indicates a minimum, and a negative second derivative indicates a maximum.

Step 4: Solve the inequality $\frac{x^2+x+2}{x^2+5 x+6}<0$. We analyze the numerator and denominator separately.

• Numerator: $N(x) = x^2 + x + 2$. The discriminant is $D = 1^2 - 4(1)(2) = 1 - 8 = -7 < 0$. Since the leading coefficient is positive, $N(x) > 0$ for all $x$.
• Denominator: $D(x) = x^2 + 5x + 6 = (x+2)(x+3)$. Since $N(x) > 0$, the inequality $\frac{N(x)}{D(x)} < 0$ is equivalent to $D(x) < 0$. So, $(x+2)(x+3) < 0$. This occurs when $-3 < x < -2$. Therefore, $x \in (-3, -2)$. Explanation: Analyzing the sign of the numerator first simplifies the problem. Then, we only need to find when the denominator is negative.

Step 5: Combine the conditions to find the range of $\lambda$. The $x$-coordinate of the local minimum, $x = -\frac{\lambda}{\sqrt{3}}$, must satisfy the inequality $-3 < x < -2$. $-3 < -\frac{\lambda}{\sqrt{3}} < -2$ Multiplying by $-\sqrt{3}$ and reversing the inequality signs: $3\sqrt{3} > \lambda > 2\sqrt{3}$ Thus, $2\sqrt{3} < \lambda < 3\sqrt{3}$. This means $\alpha = 2\sqrt{3}$ and $\beta = 3\sqrt{3}$. Explanation: We substituted the x-value of the local minimum into the solution of the inequality and solved for lambda.

Step 6: Calculate $\alpha^2 + \beta^2$. $\alpha^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12$ $\beta^2 = (3\sqrt{3})^2 = 9 \cdot 3 = 27$ $\alpha^2 + \beta^2 = 12 + 27 = 39$

3. Common Mistakes & Tips

• Sign Errors: Be extremely careful when multiplying or dividing inequalities by negative numbers. Remember to reverse the inequality signs.
• Second Derivative Test: Make sure to correctly interpret the second derivative test. $f''(x) > 0$ implies a local minimum, not a maximum.
• Checking the Numerator: Always check the sign of the numerator in a rational inequality. If it's always positive or always negative, it simplifies the problem considerably.

4. Summary

We found the local minimum of the given function using derivatives. Then, we solved the rational inequality to find the valid range for the x-coordinate of the local minimum. Finally, we combined these results to find the range for $\lambda$, which allowed us to calculate $\alpha^2 + \beta^2$.

5. Final Answer

The final answer is \boxed{39}.