Question

find the value of the derivative (if it exists) at each indicated extremum.

Answer 1

Let's begin with the derivate

$\begin{gathered} f(x)=-3x\sqrt{x+1} \\ f^{\prime}(x)=-3\left(\frac{d}{dx}\left(x\right)\sqrt{x+1}+\frac{d}{dx}\left(\sqrt{x+1}\right)x\right) \\ f^{\prime}(x)=-3\left(1\cdot\sqrt{x+1}+\frac{1}{2\sqrt{x+1}}x\right) \end{gathered}$

Simplify

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

Now let's replace the first point x=-1

$f^{\prime}(-1)=-\frac{3(3(-1)+2)}{2\sqrt{(-1)+1}}$

Solving

$\begin{gathered} f^{\prime}(-1)=-\frac{3(-3+2)}{2\sqrt{0}} \\ \\ f^{\prime}(-1)=-\frac{3(-1)}{2\cdot0} \\ \\ f^{\prime}(-1)=-\frac{-3}{0} \end{gathered}$

The answer shows us that the derivate at this point is Undefined

For x = -2/3

$\begin{gathered} f^{\prime}(-\frac{2}{3})=-\frac{3(3(-\frac{2}{3})+2)}{2\sqrt{(-\frac{2}{3})+1}} \\ \\ f^{\prime}(-\frac{2}{3})=-\frac{3(-2+2)}{2\sqrt{\frac{1}{3}}} \\ \\ f^{\prime}(-\frac{2}{3})=-\frac{3(0)}{2\sqrt{\frac{1}{3}}} \\ \\ f^{\prime}(-\frac{2}{3})=-\frac{0}{2\sqrt{\frac{1}{3}}}=0 \end{gathered}$

At this point, the derivate is 0