# Math Help - Finding domain of function

1. ## Finding domain of function

Find domain:
y= the square root of ( (4-x)^2 - 5 )

I set what was under the square root greater than or equal to zero. and came up with one answer however, I don't understand why my book tells me to put (4-x) greater than or equal to square root of 5 and (4-x) less than or equal to - square root of 5. I'm only accustomed to setting up two equations like they did with absolute values... Thanks.

2. ## Re: Finding domain of function

study the sign of the function inside the square root and exclude the interval which the function is negative since the square root of negative is undefined in Real numbers

for example square root of ( x^2 - 9 )

study x^2 - 9 sign
first find the root of it

(x-3 )( x+3 ) = 0 so x = - 3 and +3

now take numbers less than -3 and between -3 and 3, and bigger than 3 sub it in x^2 - 9
take -4, 1 , 4
-4 give us (-4)^2 - 9 = 7 > 0 positive so any number less than -3 will give us positive

1 , 1^2 - 9 < 0 so any number between -3 and 3 will give us negative
4 will give us positive now

the domain of square root of ( x^2 - 9 ) will be all real numbers except the numbers between -3 and 3

your book said that because we want the intervals where (4-x)^2 - 5 positive

(4-x)^2 - 5 >= 0

(4-x)^2 >= 5

l 4 - x l >= square 5

you should know that

$\mid x \mid \leq a \Rightarrow -a \leq x \leq a$

$a \leq \mid x \mid \Rightarrow a \leq x \;\;\; or \;\;\;x \leq -a$

3. ## Re: Finding domain of function

Shookran, Amer. So we must substitue |4 - x|^(2)for (4-x)^(2)? That makes some sense to me, since whatever is squared is going to be positive anyway. Was that what I need to be looking for with these types of problems? I just wasn't inclined to change it to absolute value.

By the way, I am also from Jordan.

4. ## Re: Finding domain of function

You have to solve $(4-x)^2-5 \geq 0 \Leftrightarrow 16-8x+x^2-5 \geq 0 \Leftrightarrow x^2-8x+11\geq 0$
You can use the quadratic formule to solve the inequality.

If you want to work with the absolut value signs (like Amer said) then:
$|4-x|\geq \sqrt{5}$
You have two possibilities:
(1) $|4-x|=4-x$ if $x<4$
Then the inequality becomes:
$4-x\geq \sqrt{5} \Leftrightarrow x\leq 4-\sqrt{5}$
(2) $|4-x|=-(4-x)=x-4$ if $x>4$
Then the inequality becomes:
$x-4\geq \sqrt{5} \Lefrightarrow x\geq \sqrt{5}+4$

So the domain is ... ?

5. ## Re: Finding domain of function

Thanks, Siron. I don't want to sound redundant, but I am still a little cloudy with the absolute value method here. Why exactly do we use the absolute value of 4 - x. Is it because the square root of five is positive or negative and therefore the left side can be posoitive or negative? I know how to solve with abs value, but I'm a tad uncomfortable with converting it to that here.

6. ## Re: Finding domain of function

Originally Posted by benny92000
Thanks, Siron. I don't want to sound redundant, but I am still a little cloudy with the absolute value method here. Why exactly do we use the absolute value of 4 - x. Is it because the square root of five is positive or negative and therefore the left side can be posoitive or negative? I know how to solve with abs value, but I'm a tad uncomfortable with converting it to that here.
we saw here how to use absolute value. that post also indicated we could solve by solving a quadratic, so let's do so, and see if we get the same answer (if this is mathematics, then by gosh we oughtta!).

$x^2-8x+11 \geq 0$

using the quadratic formula to solve when we have equality, we get:

$x = \frac{8 \pm \sqrt{20}}{2} = 4 \pm \sqrt{5}$.

now, for the product (x - a)(x - b) to be non-negative,

we either have to have: both x ≥ a and x ≥ b, -OR-

both x ≤ a, and x ≤ b.

since 4+√5 > 4-√5, for x to be ≥ both, this is the same as: x ≥ 4+√5.

for x to be LESS than both, we need x ≤ 4-√5.

so the domain of f is $(-\infty,4-\sqrt{5}] \cup [4+\sqrt{5},\infty)$

which is the same domain we found by considering absolute values.

7. ## Re: Finding domain of function

in schools we took that

$\sqrt{ x^2 } = \mid x \mid$

$x^2 = 5 \Rightarrow\;\;\; \sqrt{x^2} = \sqrt{5} \Rightarrow\;\;\; \mid x \mid = \sqrt{5} \Rightarrow\;\;\; x=\left{\sqrt{-5},\sqrt{5}\right}$

another example

$(6 -x)^2 = 9\Rightarrow\;\sqrt{(6-x)^2} = \sqrt{9}\Rightarrow\mid (6-x)\mid=3\Rightarrow\;6-x=3\;\; or \;\; x-6 = 3$

same in inequalities

$(7 - x)^2 \leq 16 \Rightarrow \sqrt{(7-x)^2} \leq \sqrt{16} \Rightarrow \mid (7-x) \mid \leq 4$

8. ## Re: Finding domain of function

That makes sense now. I never learned that in schools, Amer, but now I understand.