$${\sqrt{{\frac{\left({\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}}}$$
Well the domain is all possible values of x so lets look at this.
you cannot divide by 0 so x cannot be -2
You cannot find the square root of a neg number so
$$\frac{(-x+3)}{(x+2)}\ge0$$
now I want to get rid of the fraction but I need to know if I am multiplying by a neg or a positive
so instead of mult by (x+2) to get rid of the fraction I am going to multiply by $${\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}^{{\mathtt{2}}}$$ because I know that is positive.
$$\\(x+2)^2\times\frac{(-x+3)}{(x+2)}\ge0 \times (x+2)^2\\\\
(x+2)(-x+3)\ge0 \\\\
-(x+2)(x-3)\ge0$$
Now if I let
y=-(x+2)(x-3)
then the statement above will be true when y is positive.
Now I can see straight off that y=-(x+2)(x-3) is a concave down parabola
It is a parabola because the degree is 2 (the highest power of x is 2)
and it is concave down because the coefficient of x^2 is -1. i.e the leading coefficient is negative.
Since it is concave down, y will be positive in the middle , not at the ends.
The roots are x=-2 and x=3
so $$y\ge0$$ when $$-2\le x\le 3$$
so what do we have here.
$$x\ne -2\qquad and \qquad -2\le x \le3$$
So the domain is $$-2< x \le 3$$ this can also be written as Domain (-2,3]
this is the graph - All makes perfect sense (to me anyway)
Feel free to ask questions
$${\sqrt{{\frac{\left({\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}}}$$
Well the domain is all possible values of x so lets look at this.
you cannot divide by 0 so x cannot be -2
You cannot find the square root of a neg number so
$$\frac{(-x+3)}{(x+2)}\ge0$$
now I want to get rid of the fraction but I need to know if I am multiplying by a neg or a positive
so instead of mult by (x+2) to get rid of the fraction I am going to multiply by $${\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}^{{\mathtt{2}}}$$ because I know that is positive.
$$\\(x+2)^2\times\frac{(-x+3)}{(x+2)}\ge0 \times (x+2)^2\\\\
(x+2)(-x+3)\ge0 \\\\
-(x+2)(x-3)\ge0$$
Now if I let
y=-(x+2)(x-3)
then the statement above will be true when y is positive.
Now I can see straight off that y=-(x+2)(x-3) is a concave down parabola
It is a parabola because the degree is 2 (the highest power of x is 2)
and it is concave down because the coefficient of x^2 is -1. i.e the leading coefficient is negative.
Since it is concave down, y will be positive in the middle , not at the ends.
The roots are x=-2 and x=3
so $$y\ge0$$ when $$-2\le x\le 3$$
so what do we have here.
$$x\ne -2\qquad and \qquad -2\le x \le3$$
So the domain is $$-2< x \le 3$$ this can also be written as Domain (-2,3]
this is the graph - All makes perfect sense (to me anyway)
Feel free to ask questions