can someone explain to me how to find the domain of a function? especially the one with the square-roots and in a fraction as well as the ones that have x ^2.
i know that it has to do with determining real numbers (i hope i am not wrong) but i dunno where to begin

Question

can someone explain to me how to find the domain of a function? especially the one with the square-roots and in a fraction as well as the ones that have x ^2. 
i know that it has to do with determining real numbers (i hope i am not wrong) but i dunno where to begin

SageG · Answer

$$f(x)= \sqrt{t^2+1}$$

this is an example of a question  and

$$g(x)=\frac{ x^2+5 }{ x+2 }$$

Hero · Answer

For an expression such as $\sqrt{a}$, the domain of the function is $a \ge 0$
For an expression such as $\dfrac{a}{b}$, the domain of the function is $\mathbb{R}, b \ne 0$

SageG · Answer

thank you, can i ask a question though... in the second one, do you always just ignore the numerator  no matter what the number is?

Hero · Answer

False. If it has a square root in the numerator, for example, you still have to consider it.

Hero · Answer

$f(x) = \sqrt{\dfrac{2x}{x-2}}$ What is the domain of this for example?

SageG · Answer

okay i get it. another question please, does the addition and subtraction matter in the calculations? someone once told me that because there is an addition in the square root of a function, it can be ignored

Hero · Answer

Example of what you're referring to please.

Hero · Answer

NO such thing as ignoring without proper justification in mathematics

SageG · Answer

okay... so it was something like $$\frac{ x+4 }{ \sqrt{x-4} }$$

and so the person solved the one above and said that you cannot solve the one denominator because there is a minus in it.

im sorry for the late reply, i was looking for the question but i couldnt find it so i improvised instead

Hero · Answer

You should probably find the question.

Hero · Answer

If the question is to find the domain, then the domain will be $\sqrt{x-4} > 0$

Hero · Answer

From there you solve for $x$

Hero · Answer

@SageG any questions concerning that?

SageG · Answer

alright i will seach for it.    and yes i have a question, regarding solving x, is the first step separating the square root into two parts, i.e. square root x minus square root of 4 ?

Hero · Answer

There's no such rule as what you are describing. You're asking if
$\sqrt{x - a} = \sqrt{x} - \sqrt{a}$ .

Where have you ever seen a rule such as that?

SageG · Answer

i havent, my sister did that and said she isnt sure so i should ask

Hero · Answer

To solve $\sqrt{x - 4} > 0$ , you begin by squaring both sides.

Hero · Answer

$(\sqrt{x - 4})^2 > 0^2$

SageG · Answer

why?

Hero · Answer

Because the goal is to solve for $x$

Hero · Answer

In order to solve for $x$ you can only use the rules of mathematics to do so.

Hero · Answer

In order to isolate $x$ in this case, you must get rid of the square root. The only way to do that is to perform the inverse operation and apply it to both sides of the equation. The inverse of square root is square.

SageG · Answer

okay so, i understand. then x-4>0 which in turn will become x>4

Hero · Answer

Correct

SageG · Answer

oh okay, thank goodness.

okay so back to your question 
    f(x) = square root 2x / x-2

x-2 $$x-2\neq0 $$  that is to say as long as x is not 2, the function has a defined value.

then$$2x \ge0$$

....is that wrong?

TETSXPREME · Answer

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