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√(x+1) \\ f^(\prime)(x)=-3\left((d)/(dx)\left(x\right)√(x+1)+(d)/(dx)\left(√(x+1)\right)x\right) \\ f^(\prime)(x)=-3\left(1\cdot√(x+1)+(1)/(2√(x+1))x\right) \end{gathered}`

Simplify

`f^(\prime)(x)=-(3\left(3x+2\right))/(2√(x+1))`

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

`f^(\prime)(-1)=-(3(3(-1)+2))/(2√((-1)+1))`

Solving

`\begin{gathered} f^(\prime)(-1)=-(3(-3+2))/(2√(0)) \\ \\ f^(\prime)(-1)=-(3(-1))/(2\cdot0) \\ \\ f^(\prime)(-1)=-(-3)/(0) \end{gathered}`

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

For x = -2/3

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

At this point, the derivate is 0