Question

The function $f$ defined by $f\left(x\right)=x^3-6x^2-36x+7$ is increasing if

• A. $x>2$ and also $x>6$.
• B. $x>2$ and also $x<6$.
• C. $x>-2$ and also $x<6$.
• D. $x<-2$ and also $x>6$.

Answer 1

The correct option is D

$x<-2$ and also $x>6$.

Explanation for the correct option

Given function, $f\left(x\right)=x^3-6x^2-36x+7$.

Differentiate the given function with respect to $x$.

$\begin{array}{rcl}\frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]& =& \frac{d}{dx}\left(x^3-6x^2-36x+7\right)\\ & =& \frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(-6x^2\right)+\frac{d}{dx}\left(-36x\right)+\frac{d}{dx}\left(7\right)\left[\because \frac{d}{d\mathrm{x}}(\mathrm{u}\pm \mathrm{v})=\frac{d\mathrm{u}}{d\mathrm{x}}\pm \frac{d\mathrm{v}}{d\mathrm{x}}\right]\\ & =& \left(3x^2\right)-\left(6\times 2x\right)-36+0\\ & =& 3x^2-12x-36\\ & =& 3\left(x^2-4x-12\right)\\ & =& 3\left(x^2-6x+2x-12\right)\\ & =& 3\left[x\left(x-6\right)+2\left(x-6\right)\right]\\ & =& 3\left(x-6\right)\left(x+2\right)\end{array}$

Now, for increasing intervals $f\text{'}\left(\mathrm{x}\right)>0$.

$$\begin{gathered} \Rightarrow 3\left(x-6\right)\left(x+2\right)>0 \\ \Rightarrow \left(x-6\right)\left(x+2\right)>0 \\ \Rightarrow x\in \left(-\infty ,-2\right)\cup \left(6,\infty \right) \end{gathered}$$

Therefore, the function $f\left(x\right)=x^3-6x^2-36x+7$ is increasing for $x<-2$ and also $x>6$.

Hence, the function is increasing if $x<-2$ and also $x>6$ and therefore correct option is (D).