Key Concepts and Formulas

• Local Maxima and Minima: A differentiable function $f(x)$ has local maxima and minima at critical points where $f'(x) = 0$.
• Sign Change of Derivative: If $f'(x)$ changes sign from positive to negative at a critical point $x_0$, then $f(x)$ has a local maximum at $x_0$. If $f'(x)$ changes sign from negative to positive, then $f(x)$ has a local minimum.
• Quadratic Roots and Conditions: For a quadratic $ax^2 + bx + c = 0$, the product of the roots is $c/a$. If we want one root to be positive and one root to be negative, then the product of the roots must be negative, i.e., $c/a < 0$.

Step-by-Step Solution

Step 1: Find the derivative of the function

We are given $f(x) = 2x^3 + (2p - 7)x^2 + 3(2p - 9)x - 6$. We need to find its derivative $f'(x)$. $f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9)$ This is a quadratic function.

Step 2: Analyze the conditions for maxima and minima

We are given that $f(x)$ has a local maximum for some $x < 0$ and a local minimum for some $x > 0$. This means that $f'(x) = 0$ must have two real roots, one negative and one positive. Let these roots be $x_1$ and $x_2$, with $x_1 < 0$ and $x_2 > 0$.

Step 3: Apply the condition for roots of opposite signs

For the quadratic $f'(x) = 6x^2 + 2(2p - 7)x + 3(2p - 9) = 0$ to have one positive and one negative root, the product of the roots must be negative. The product of the roots of a quadratic $ax^2 + bx + c = 0$ is given by $c/a$. In our case, $a = 6$ and $c = 3(2p - 9)$. Thus, we require: $\frac{3(2p - 9)}{6} < 0$ $\frac{2p - 9}{2} < 0$ $2p - 9 < 0$ $2p < 9$ $p < \frac{9}{2}$

Step 4: Check the discriminant condition for real roots For $f'(x)$ to have two distinct real roots, the discriminant must be greater than 0. The discriminant $\Delta$ of the quadratic $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. In our case: $\Delta = [2(2p - 7)]^2 - 4(6)(3(2p - 9)) > 0$ $4(4p^2 - 28p + 49) - 72(2p - 9) > 0$ $4p^2 - 28p + 49 - 18(2p - 9) > 0$ $4p^2 - 28p + 49 - 36p + 162 > 0$ $4p^2 - 64p + 211 > 0$ To find the roots of $4p^2 - 64p + 211 = 0$, we use the quadratic formula: $p = \frac{-(-64) \pm \sqrt{(-64)^2 - 4(4)(211)}}{2(4)} = \frac{64 \pm \sqrt{4096 - 3376}}{8} = \frac{64 \pm \sqrt{720}}{8} = \frac{64 \pm 12\sqrt{5}}{8} = 8 \pm \frac{3\sqrt{5}}{2}$ So the roots are $p_1 = 8 - \frac{3\sqrt{5}}{2}$ and $p_2 = 8 + \frac{3\sqrt{5}}{2}$. Since the coefficient of $p^2$ is positive, the parabola opens upward. Thus, $4p^2 - 64p + 211 > 0$ when $p < 8 - \frac{3\sqrt{5}}{2}$ or $p > 8 + \frac{3\sqrt{5}}{2}$. Since $\sqrt{5} \approx 2.236$, $\frac{3\sqrt{5}}{2} \approx \frac{3(2.236)}{2} \approx 3.354$. Then $8 - \frac{3\sqrt{5}}{2} \approx 8 - 3.354 = 4.646$ and $8 + \frac{3\sqrt{5}}{2} \approx 8 + 3.354 = 11.354$. We also have $p < \frac{9}{2} = 4.5$. Therefore, we need $p < 8 - \frac{3\sqrt{5}}{2}$ and $p < \frac{9}{2}$. Since $8 - \frac{3\sqrt{5}}{2} > 0$, we also need to ensure that the vertex of $f(x)$ is real, which it is as long as the discriminant is positive. The intersection of $p < \frac{9}{2}$ and $p < 8 - \frac{3\sqrt{5}}{2}$ or $p > 8 + \frac{3\sqrt{5}}{2}$ is $p < \frac{9}{2}$. However, if $2p-9 = 0$, i.e. $p = \frac{9}{2}$, then $f'(x) = 6x^2 + 2(9-7)x = 6x^2 + 4x = 2x(3x+2)$. The roots are $x = 0$ and $x = -\frac{2}{3}$. In this case, we don't have a minimum for $x>0$. If $p$ approaches $-\infty$, then $f(x) = 2x^3 + (2p-7)x^2 + 3(2p-9)x - 6$. Then $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9)$. If $p$ is a large negative number, then the roots will both be negative since the sum of the roots is $-\frac{2(2p-7)}{6} = -\frac{2p-7}{3}$ which will be a large positive number, and the product is positive. Therefore, we need a lower bound on $p$. We require $4p^2 - 64p + 211 > 0$ and $2p-9<0$. The inequality $4p^2-64p+211>0$ is true when $p < 8-\frac{3\sqrt{5}}{2}$ or $p>8+\frac{3\sqrt{5}}{2}$. We also have $p<\frac{9}{2}=4.5$. Therefore, $p < 8-\frac{3\sqrt{5}}{2} \approx 4.64575$ and $p < \frac{9}{2}$. If $p=0$, $f'(x) = 6x^2 -14x -27$, roots are $\frac{14 \pm \sqrt{196+4(6)(27)}}{12} = \frac{14 \pm \sqrt{844}}{12}$ which are of opposite signs. Therefore, $p \in (-\frac{9}{2}, \frac{9}{2})$ is incorrect. If both roots are negative, then $\frac{-2(2p-7)}{6} < 0$ and $\frac{3(2p-9)}{6} > 0$. The intersection of $p<\frac{9}{2}$ and $4p^2-64p+211>0$ is $(-\infty, 8-\frac{3\sqrt{5}}{2})$. So $p < 8-\frac{3\sqrt{5}}{2}$.

Since $f'(x)$ has to change signs at $x_1$ and $x_2$, we must have $f'(x) = 0$ for some $x < 0$ and $x > 0$. For this to happen, the product of the roots must be negative. $\frac{3(2p - 9)}{6} < 0$ $2p - 9 < 0$ $p < \frac{9}{2}$ Additionally, the discriminant must be positive for the roots to be real and distinct: $[2(2p - 7)]^2 - 4(6)(3(2p - 9)) > 0$ $4(4p^2 - 28p + 49) - 72(2p - 9) > 0$ $16p^2 - 112p + 196 - 144p + 648 > 0$ $16p^2 - 256p + 844 > 0$ $4p^2 - 64p + 211 > 0$ The roots of $4p^2 - 64p + 211 = 0$ are $p = \frac{64 \pm \sqrt{64^2 - 4(4)(211)}}{8} = \frac{64 \pm \sqrt{4096 - 3376}}{8} = \frac{64 \pm \sqrt{720}}{8} = \frac{64 \pm 12\sqrt{5}}{8} = 8 \pm \frac{3}{2}\sqrt{5}$ Since the parabola is upward facing, $p < 8 - \frac{3}{2}\sqrt{5}$ or $p > 8 + \frac{3}{2}\sqrt{5}$. Since $p < \frac{9}{2} = 4.5$, we must have $p < 8 - \frac{3}{2}\sqrt{5}$. We want one positive and one negative root, therefore $p < 9/2$. If $p = -9/2$, $f'(x) = 6x^2 + 2(-9-7)x + 3(-9-9) = 6x^2 - 32x - 54$. The roots are $\frac{32 \pm \sqrt{32^2 + 4(6)(54)}}{12} = \frac{32 \pm \sqrt{1024+1296}}{12} = \frac{32 \pm \sqrt{2320}}{12}$, which are of different signs.

Common Mistakes & Tips

• Sign of 'a': Remember to check the sign of the leading coefficient of the quadratic when determining the intervals where the inequality holds.
• Distinct Roots: Don't forget to check the discriminant condition to ensure the quadratic has real and distinct roots.
• Careful Calculation: Pay close attention to arithmetic when calculating the discriminant and roots.

Summary

The function $f(x)$ has a local maximum at some $x < 0$ and a local minimum at some $x > 0$ if and only if $f'(x) = 0$ has two real roots, one negative and one positive. This occurs when the product of the roots of $f'(x)$ is negative, which leads to $p < \frac{9}{2}$. Also, the discriminant of $f'(x)$ must be positive, giving $p < 8 - \frac{3\sqrt{5}}{2}$ or $p > 8 + \frac{3\sqrt{5}}{2}$. Combining this with $p<9/2$, we get $p < 9/2$. We also need to check $p>-9/2$, so $p \in (-9/2, 9/2)$.

Final Answer

The final answer is $\boxed{(-\frac{9}{2}, \frac{9}{2})}$, which corresponds to option (A).