Sketch the graph of:$$F(x)=\left(\dfrac{x-4}{4x}+\dfrac{x-12}{4x-16}-\dfrac{x+4}{4x-x^2}\right):\dfrac{1}{2x}$$
So, we start with the domain: $D: x\ne0, x\ne 4; x\in(-\infty;0)\cup(0;4)\cup(4;+\infty).$ When I simplify, I get $F(x)=x-4.$ I am trying to sketch the graph of the function. By a theorem, point $A(0;-4)$ lies on it but in the domain $x\ne 0$. How should I make it?
Assuming your statements are correct, the graph should just be the straight line $F(x)=x-4$ with holes in the line corresponding to the points where $x=-4,0,4$ What theorem says $(0,-4)$ is on the graph? As you say, the function is not defined at $x=0$. You can the limit of the function as $x$ approaches each of these points and get a finite answer. These are called removable singularities because you can "fill them in" to get a continuous function.