91Ó°ÊÓ

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.

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 :-

localid="1649084613709" $$\begin{gathered} f\text{'}\left(x\right)=\left(x^2+3\right)\left[\frac{3}{2}{\left(x-2\right)}^{\frac{3}{2}-1}\right]+{\left(x-2\right)}^{\frac{3}{2}}\left(2x\right) \\ ⇒f\text{'}\left(x\right)=\left(x^2+3\right)\left(\frac{3}{2}{\left(x-2\right)}^{\frac{1}{2}}\right)+2x{\left(x-2\right)}^{\frac{3}{2}} \\ ⇒f\text{'}\left(x\right)=\frac{3}{2}\left(x^2+3\right){\left(x-2\right)}^{\frac{1}{2}}+2x{\left(x-2\right)}^{\frac{3}{2}} \\ ⇒f\text{'}\left(x\right)={\left(x-2\right)}^{\frac{1}{2}}\left[\frac{3}{2}\left(x^2+3\right)+2x\left(x-2\right)\right] \\ ⇒f\text{'}\left(x\right)={\left(x-2\right)}^{\frac{1}{2}}\left[\frac{3x^2+9+4x^2-8x}{2}\right] \\ ⇒f\text{'}\left(x\right)=\sqrt{\left(x-2\right)}\left[\frac{7x^2-8x+9}{2}\right] \end{gathered}$$

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 :-

role="math" localid="1649084666498" $$\begin{gathered} \sqrt{\left(x-2\right)}\left[\frac{7x^2-8x+9}{2}\right]=0 \\ ⇒\sqrt{\left(x-2\right)}\left(7x^2-8x+9\right)=0 \\ ⇒\sqrt{x-2}=0and7x^2-8x+9=0 \\ ⇒x-2=0and7x^2-8x+9=0 \\ ⇒x=2and7x^2-8x+9=0 \end{gathered}$$

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$x=2$.

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.