Question:

Step 1: Find the first derivative of the function. To find the points of local maxima and minima, we first need to compute the derivative of the given function $f(x)$. The function is $f(x)=2x^3+(2p-7)x^2+3(2p-9)x-6$. $$f'(x) = \frac{d}{dx}\left(2x^3+(2p-7)x^2+3(2p-9)x-6\right)$$ $$f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9)$$ Step 2: Relate the conditions on maxima and minima to the roots of $f'(x)=0$. The critical points, where local maxima or minima can occur, are the roots of the equation $f'(x)=0$. The problem states that $f(x)$ has a maxima for some value of $x<0$ and a minima for some value of $x>0$. This implies that $f'(x)=0$ must have two distinct real roots, one of which is negative and the other is positive. Let the roots of $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9) = 0$ be $x_1$ and $x_2$. We require $x_1 < 0$ and $x_2 > 0$. Step 3: Apply conditions for roots of a quadratic equation. For a quadratic equation $Ax^2 + Bx + C = 0$ to have two distinct real roots of opposite signs, the product of the roots must be negative ($C/A < 0$). This condition automatically implies that the discriminant $D = B^2 - 4AC$ is positive, ensuring distinct real roots. In our case, $f'(x) = 6x^2 + 2(2p-7)x + 3(2p-9) = 0$, so we have: $A = 6$ $B = 2(2p-7)$ $C = 3(2p-9)$ The product of the roots is $\frac{C}{A}$. We must have: $$\frac{3(2p-9)}{6} < 0$$ $$\frac{2p-9}{2} < 0$$ Multiplying by 2, we get: $$2p-9 < 0$$ $$2p < 9$$ $$p < \frac{9}{2}$$ Step 4: Confirm the nature of critical points. Since the leading coefficient of $f'(x)$ is $6 > 0$, the parabola $y = f'(x)$ opens upwards. If $f'(x)$ has roots $x_1 < 0$ and $x_2 > 0$, then: - For $x < x_1$, $f'(x) > 0$, so $f(x)$ is increasing. - For $x_1 < x < x_2$, $f'(x) < 0$, so $f(x)$ is decreasing. - For $x > x_2$, $f'(x) > 0$, so $f(x)$ is increasing. This means $f(x)$ has a local maximum at $x_1$ (where $x_1 < 0$) and a local minimum at $x_2$ (where $x_2 > 0$). This precisely matches the conditions given in the problem statement. Step 5: State the final set of values for $p$. Based on the analysis, the set of all values of $p$ for which the given conditions are satisfied is $p < \frac{9}{2}$. In interval notation, this is $p \in \left(-\infty, \frac{9}{2}\right)$. The final answer is $\boxed{\left(-\infty,\dfrac{9}{2}\right)}$.