Answer
$F(x)$ has an absolute maximum value of $1$ at $x=-1$ and an absolute minimum value of $1/2$ at $x=-2$ on the defined domain.
Work Step by Step
$$F(x)=-\frac{1}{x}\hspace{1cm}-2\le x\le-1$$
1) Find the derivative $F'(x)$: $$F'(x)=-\frac{(1)'\times x-1\times(x)'}{x^2}=-\frac{0\times x-1}{x^2}$$ $$F'(x)=-\frac{(-1)}{x^2}=\frac{1}{x^2}$$
Here there is no value of $x$ for which $F'(x)=0$. However, we can see that $F'(x)$ is undefined at $x=0$, which means $x=0$ is a critical point of $F(x)$.
However, since the defined domain here is $[-2,-1]$, we would not take $x=0$ into consideration.
2) Evaluate $F(x)$ at the endpoints:
- For $x=-2$: $$F(-2)=-\frac{1}{-2}=\frac{1}{2}$$
- For $x=-1$: $$F(-1)=-\frac{1}{-1}=1$$
This means the function $F(x)$ has an absolute maximum value of $1$ at $x=-1$ and an absolute minimum value of $1/2$ at $x=-2$ on the defined domain.
