Q 1 - Calculus

Question

Finding critical points: For each of the following functions $f$, find all of the $x$-values for which $f\text{'}\left(x\right)=0$ and all of the $x$-values for which $f\text{'}\left(x\right)$ does not exist.

$f\left(x\right)=x^3\sqrt{3x+1}$

Step-by-Step Solution

Answer

The points for which $f\text{'}\left(x\right)=0$ are :-

$0 and -\frac{2}{7}$

The point for which $f\text{'}\left(x\right)$ does not exist is :-

$-\frac{1}{3}$

Step 1. Given Information

We have given the following function :-

$f\left(x\right)=x^3\sqrt{3x+1}$

We have to find the points for which $f\text{'}\left(x\right)=0$.

Also we have to find the points for which $f\text{'}\left(x\right)$ does not exist.

Firstly we will find the derivative, then we will find the required points.

Step 2. Find the derivative of the given function

The given function is :-

$f\left(x\right)=x^3\sqrt{3x+1}$

Use product rule of derivative to differentiate this function, then we have :-

$f\text{'}\left(x\right)=x^3\frac{d}{dx}\left(\sqrt{3x+1}\right)+\sqrt{3x+1}\frac{d}{dx}\left(x^3\right)$

Use power rule :-

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=x^3\frac{d}{dx}{(3x+1)}^{\frac{1}{2}}+\sqrt{3x+1}\left(3x^2\right) \\ \Rightarrow f\text{'}\left(x\right)=x^3\left[\frac{1}{2}{(3x+1)}^{\frac{1}{2}-1}\times 3\right]+3x^2\sqrt{3x+1} \\ \Rightarrow f\text{'}\left(x\right)=x^3\left[\frac{1}{2}{(3x+1)}^{-\frac{1}{2}}\times 3\right]+3x^2\sqrt{3x+1} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3x^3}{2{(3x+1)}^{\frac{1}{2}}}+3x^2\sqrt{3x+1} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3x^3}{2\sqrt{3x+1}}+3x^2\sqrt{3x+1} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3x^3+6x^2\sqrt{3x+1}\sqrt{3x+1}}{2\sqrt{3x+1}} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3x^3+6x^2(3x+1)}{2\sqrt{3x+1}} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3x^3+18x^3+6x^2}{2\sqrt{3x+1}} \\ \Rightarrow f\text{'}\left(x\right)=\frac{21x^3+6x^2}{2\sqrt{3x+1}} \end{gathered}$$

Step 3. Put f ' ( x ) = 0

We find that :-

$f\text{'}\left(x\right)=\frac{21x^3+6x^2}{2\sqrt{3x+1}}$

Now put $f\text{'}\left(x\right)=0$, then we have :-

$$\begin{gathered} \frac{21x^3+6x^2}{2\sqrt{3x+1}}=0 \\ \Rightarrow 21x^3+6x^2=0 \\ \Rightarrow 3x^2(7x+2)=0 \\ \Rightarrow 3x^2=0 and 7x+2=0 \\ \Rightarrow x^2=0 and 7x=-2 \\ \Rightarrow x=0 and x=-\frac{2}{7} \end{gathered}$$

The values of $x$ for which $f\text{'}\left(x\right)=0$ are $0 and -\frac{2}{7}$.

Step 4. The values for which f ' ( x ) does not exist

We find that :-

$f\text{'}\left(x\right)=\frac{21x^3+6x^2}{2\sqrt{3x+1}}$

We know that a function does not exist where the denominator is equals to zero, then we have :-

$$\begin{gathered} 2\sqrt{3x+1}=0 \\ \Rightarrow \sqrt{3x+1}=0 \\ \Rightarrow 3x+1=0 \\ \Rightarrow 3x=-1 \\ \Rightarrow x=-\frac{1}{3} \end{gathered}$$

So the value of $x$ for which $f\text{'}\left(x\right)$ does not exist is $-\frac{1}{3}$.