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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(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 $p = \dots$

Similar Questions

The interval in which the function $f(x) = 2x^2 - \log x$,for $x > 0$ decreases is

In which interval is the function $f(x) = \tan^{-1}(\sin x + \cos x)$ an increasing function?

If $y = ax^3 + 3x^2 + (2a + 1)x + 1000$ is a strictly increasing function for all values of $x$,then:

If $f(x)=e^{x}(x-2)^{2}$, then

The function $f(x) = x^x$ is a decreasing function in the interval of . . . . . . . $(x \in R^{+})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module