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 $x>6$.

The explanation for the correct option

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

Differentiate the given function with respect to $x$.

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

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

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

Hence, the correct option is $\mathrm{Option}\left(\mathrm{D}\right)$