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# what is the domain of this function ? sqrt((-x+3)/(x+2))

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what is the domain of this function ? sqrt((-x+3)/(x+2))

tnx

Guest Jan 2, 2015

#1
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$${\sqrt{{\frac{\left({\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}}}$$

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)

Melody  Jan 2, 2015
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#1
+91900
+10

$${\sqrt{{\frac{\left({\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}}}$$

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)