Let f(x) = x^3 + 6x^2 + px + 2. If f(x) is a | vedclass

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(B) Given the function $f(x) = x^3 + 6x^2 + px + 2$.To find the interval where the function is decreasing,we calculate the derivative:$f'(x) = 3x^2 +

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$9$

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(B) Given the function $f(x) = x^3 + 6x^2 + px + 2$.To find the interval where the function is decreasing,we calculate the derivative:$f'(x) = 3x^2 + 12x + p$.For $f(x)$ to be a decreasing function on the interval $(-3, -1)$,we must have $f'(x) \leq 0$ for all $x \in (-3, -1)$.The quadratic expression $3x^2 + 12x + p$ must be less than or equal to $0$ between its roots $-3$ and $-1$.Thus,the roots of the equation $3x^2 + 12x + p = 0$ are $x = -3$ and $x = -1$.Using the product of roots formula for a quadratic equation $ax^2 + bx + c = 0$,the product of roots is $\frac{c}{a}$.Therefore,$(-3) 	imes (-1) = \frac{p}{3}$.$3 = \frac{p}{3} \implies p = 9$.

Let $f(x) = x^3 + 6x^2 + px + 2$. If $f(x)$ is a decreasing function on the interval $(-3, -1)$,then $p = \dots$

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