Q 10 - 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.

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

Step-by-Step Solution

Answer

There exist no $x$-value for which $f\text{'}\left(x\right)=0$.

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 :-

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

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 :-

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

We can write it as :-

$$\begin{gathered} f\left(x\right)=\frac{1}{\sqrt{1+3x}} \\ \Rightarrow f\left(x\right)=\frac{1}{{(1+3x)}^{{\displaystyle \frac{1}{2}}}} \\ \Rightarrow f\left(x\right)={(1+3x)}^{-\frac{1}{2}} \end{gathered}$$

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

$$\begin{gathered} f\text{'}\left(x\right)={\frac{d}{dx}(1+3x)}^{-\frac{1}{2}} \\ \Rightarrow f\text{'}\left(x\right)=-\frac{1}{2}{(1+3x)}^{-\frac{1}{2}-1} \\ \Rightarrow f\text{'}\left(x\right)=-\frac{1}{2}{(1+3x)}^{-\frac{3}{2}} \\ \Rightarrow f\text{'}\left(x\right)=-\frac{1}{2{(1+3x)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$ 

Step 3. Put f ' ( x ) = 0

We find that :-

$f\text{'}\left(x\right)=-\frac{1}{2{(1+3x)}^{{\displaystyle \frac{3}{2}}}}$

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

$$\begin{gathered} -\frac{1}{2{(1+3x)}^{{\displaystyle \frac{3}{2}}}}=0 \\ \Rightarrow -1=0 \end{gathered}$$

This is not possible.

That is there exist no $x$-value for which $f\text{'}\left(x\right)=0$.

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

We find that :-

$f\text{'}\left(x\right)=-\frac{1}{2{(1+3x)}^{{\displaystyle \frac{3}{2}}}}$

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

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

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