Q 3 - 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.

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

Step-by-Step Solution

Answer

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

$4 and \frac{1}{4}$.

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

$0 and 1$.

Step 1. Given Information

We have given the following function :-

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

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 :-

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

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

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

Use product rule and power rule to find derivative, then we have :-

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

Step 3. Put f ' ( x ) = 0

We find that :-

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

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

$$\begin{gathered} \frac{-\left(\sqrt{x}+2x+2\right)}{2x^2{(x-1)}^{\frac{3}{2}}}=0 \\ \Rightarrow -\left(\sqrt{x}+2x+2\right)=0 \\ \Rightarrow \sqrt{x}+2x+2=0 \\ \Rightarrow \sqrt{x}+2x=-2 \\ \Rightarrow \sqrt{x}\left(1+2\sqrt{x}\right)=2 \\ \Rightarrow \sqrt{x}=2 and 1+2\sqrt{x}=2 \\ \Rightarrow x=2^2 and 2\sqrt{x}=1 \\ \Rightarrow x=4 and \sqrt{x}=\frac{1}{2} \\ \Rightarrow x=4 and x={\left(\frac{1}{2}\right)}^2 \\ \Rightarrow x=4 and x=\frac{1}{4} \end{gathered}$$

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

We find that :- 

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

We know that a function does not exist where the denominator is equals to zero.  Then we have :-
$$\begin{gathered} 2x^2{(x-1)}^{\frac{3}{2}}=0\Rightarrow 2x^2=0 and {(x-1)}^{\frac{3}{2}}=0 \\ \Rightarrow x^2=0 and (x-1)=0 \\ \Rightarrow x=0 and x=1 \end{gathered}$$

That is $f\text{'}\left(x\right)$ is not defines for points $x=0$ and $x=1$.