Question

The which interval the given function $f\left(x\right)=-2x^3-9x^2-12x+1$ is decreasing

• A. $\left(-2,\infty \right)$
• B. $\left[-2,-1\right]$
• C. $\left(-\infty ,-1\right)$
• D. $\left(-\infty ,-2\right)$ and $\left(-1,-\infty \right)$

Answer 1

The correct option is D

$\left(-\infty ,-2\right)$ and $\left(-1,-\infty \right)$

Explanation of the correct option:

Given : $f\left(x\right)=-2x^3-9x^2-12x+1$

Differentiate with respect to $x$,

$\Rightarrow$ $f\text{'}\left(x\right)=-6x^2-18x-12$

$\Rightarrow$ $f\text{'}\left(x\right)=-6\left(x+2\right)\left(x+1\right)$

To find the point where slope is changing put $f\text{'}\left(x\right)=0$,

$\Rightarrow 6\left(x+2\right)(x+1)=0$

Thus, at $x=-2$ and $x=-1$ slope is changing.

We know that for $f\text{'}\left(x\right)>0$ function is increasing and for $f\text{'}\left(x\right)<0$ function is decreasing.

Interval	Value of $x$	Value of $f\text{'}\left(x\right)$	Increasing or decreasing
$\left(-\infty ,-2\right)$	$x=-3$	$\begin{array}{rcl}f\text{'}(-3)& =& -6(-3+2)(-3+1)\\ & =& -6(-1)(-2)\\ & =& -12\end{array}$	Since, $f\text{'}\left(x\right)<0$ function is decreasing.
$\left(-2,-1\right)$	$x=-1.5$	$\begin{array}{rcl}f\text{'}(-1.5)& =& -6(-1.5+2)(-1.5+1)\\ & =& 6(0.5)(-0.5)\\ & =& +1.5\end{array}$	Since, $f\text{'}\left(x\right)>0$ function is increasing.
$\left(-1,\infty \right)$	$x=0$	$\begin{array}{rcl}f\text{'}(-1.5)& =& -6(0+2)(0+1)\\ & =& -6\left(2\right)\left(1\right)\\ & =& -12\end{array}$	Since, $f\text{'}\left(x\right)<0$ function is decreasing.

Since, $f\text{'}\left(x\right)<0$ for $\left(-\infty ,-2\right)$ and $\left(-1,-\infty \right)$.

Therefore $f\left(x\right)$ is decreasing in $\left(-\infty ,-2\right)$ and $\left(-1,-\infty \right)$

Hence, option (D) is the correct option.