Q 6 - 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.

$f\left(x\right)={\left(x\sqrt{x+1}\right)}^{-2}$.

Step-by-Step Solution

Answer

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

$x=-\frac{2}{3}$.

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

$x=0 and x=-1$.

Step 1. Given Information

We have given the following function :-

$f\left(x\right)={\left(x\sqrt{x+1}\right)}^{-2}$.

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 :-

$f\left(x\right)={\left(x\sqrt{x+1}\right)}^{-2}$

Use power rule and chain rule to differentiate this function, then we have :-

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

Now use product rule :-

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

Simplify it :-

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

Step 3. Put f ' ( x ) = 0

We find that :-

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

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

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

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

We find that :-

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

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

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