Q 2 - 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.

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

Step-by-Step Solution

Answer

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

$x=2$.

There is no point for which $f\text{'}\left(x\right)$ does not exist.

Step 1. Given Information

We have given the following function :- 

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

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 :-

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

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

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

Use power rule and chain rule :-

$$\begin{gathered} f\text{'}\left(x\right)=\left(x^2+3\right)\left[\frac{3}{2}{\left(x-2\right)}^{\frac{3}{2}-1}\right]+{\left(x-2\right)}^{\frac{3}{2}}\left(2x\right) \\ \Rightarrow f\text{'}\left(x\right)=\left(x^2+3\right)\left(\frac{3}{2}{\left(x-2\right)}^{\frac{1}{2}}\right)+2x{\left(x-2\right)}^{\frac{3}{2}} \\ \Rightarrow f\text{'}\left(x\right)=\frac{3}{2}\left(x^2+3\right){\left(x-2\right)}^{\frac{1}{2}}+2x{\left(x-2\right)}^{\frac{3}{2}} \\ \Rightarrow f\text{'}\left(x\right)={\left(x-2\right)}^{\frac{1}{2}}\left[\frac{3}{2}\left(x^2+3\right)+2x\left(x-2\right)\right] \\ \Rightarrow f\text{'}\left(x\right)={\left(x-2\right)}^{\frac{1}{2}}\left[\frac{3x^2+9+4x^2-8x}{2}\right] \\ \Rightarrow f\text{'}\left(x\right)=\sqrt{\left(x-2\right)}\left[\frac{7x^2-8x+9}{2}\right] \end{gathered}$$

Step 3. Put f ' ( x ) = 0

We find that :- 

$f\text{'}\left(x\right)=\sqrt{\left(x-2\right)}\left[\frac{7x^2-8x+9}{2}\right]$

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

$$\begin{gathered} \sqrt{\left(x-2\right)}\left[\frac{7x^2-8x+9}{2}\right]=0 \\ \Rightarrow \sqrt{\left(x-2\right)}\left(7x^2-8x+9\right)=0 \\ \Rightarrow \sqrt{x-2}=0 and 7x^2-8x+9=0 \\ \Rightarrow x-2=0 and 7x^2-8x+9=0 \\ \Rightarrow x=2 and 7x^2-8x+9=0 \end{gathered}$$

The roots of the equation $7x^2-8x+9=0$ are imaginary.

So the $x$-value for which $f\text{'}\left(x\right)=0$ is $x=2$.

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

We find that :-

$f\text{'}\left(x\right)=\sqrt{\left(x-2\right)}\left[\frac{\left(7x^2-8x+9\right)}{2}\right]$

We know that a function does not exist where the denominator is equals to zero.

We can see that the denominator of the function is constant.

So it cannot be zero.

So there exist no point for which $f\text{'}\left(x\right)$ does not exist.