how to find domain of this?

Question

anonymous · Answer

$$y=1/(\sqrt(1-(x/2))+1)$$

anonymous · Answer

the +1 is not in sqrt

e.mccormick · Answer

Know what restrictons can exist on the domain?

e.mccormick · Answer

As in, in general

anonymous · Answer

higher than 0

anonymous · Answer

i confused about the process

anonymous · Answer

how do i get the domain, what do i equate to what

e.mccormick · Answer

When dealing with reals, there can not be a negative under an even root.  In all cases, the denominator of a fraction can not be 0.

e.mccormick · Answer

So what you do is find out when those things would happen.

anonymous · Answer

the answer must be greater than 0

anonymous · Answer

$$y=\frac{ 1 }{ (\sqrt1-(\frac{ x }{ 2 }) + 1 }$$

anonymous · Answer

this is the real equation

anonymous · Answer

the x/2 is UNDER the square root

e.mccormick · Answer

The answer can be lots of things.  It does not have to be greater than 0.

$\dfrac{1}{\sqrt{1-\frac{x}{2}}+1}$

or

$\dfrac{1}{\sqrt{1}-\frac{x}{2}+1}$

Your (( ) is confusing.

e.mccormick · Answer

Ah.

anonymous · Answer

first one

e.mccormick · Answer

OK, so the entire fraction, the bottom can not be 0.  That is one domain restriction.

The root, because it is even, can not be negative under there.  That is a second restriction.

Find out when those two things happen.  Then you can write a domain that avoids thosr problems.

anonymous · Answer

what can equate to start off?

e.mccormick · Answer

See the entire bottom of the fraction?  Take that whole thing and set it to 0.  Then solve for x.  When that is 0, it is invalid.  You need to test also to make sure how you deal with the root does not cause or loose any answers.

e.mccormick · Answer

Then set the radical to <0 and solve that.  Again, any time it is that, it is invalid.

anonymous · Answer

i got x = -4

anonymous · Answer

for the second one i got x > 2

anonymous · Answer

oh the second one is x <2

e.mccormick · Answer

I see the < 2.  -4, hmmm...

anonymous · Answer

that one is x = 0

anonymous · Answer

sorry -4 is wrong

anonymous · Answer

x=0

e.mccormick · Answer

Yah, which is due to the $\pm$ thing.  If you think about it logically, what would make it 0 on the bottom is when what is under the root evaluated to -1.

e.mccormick · Answer

$\sqrt{1-\frac{x}{2}}+1=0$

Well, -1 +1 =0

$ \sqrt{1-\frac{x}{2}}=-1$

So that is, well, either never or due to the $\pm$ thing with square roots when it is 1.  Well, for it to be 1, the fraction would have to be 0.

That is just using some logic.  So there may be  hole at x=0, and it is certainly never x > 2.

anonymous · Answer

how would you go by finding the domain. can you please list your steps?

e.mccormick · Answer

By knowing what is invalid as an input, you find the domain.  It is everything else.

Did you test what you got because of the square root?  I warned about that.  Try putting in somethin above 2 and try 0.  If one or both gets you invalid situations, that is a restriction.

e.mccormick · Answer

$f(x)=\dfrac{1}{\sqrt{1-\frac{x}{2}}+1}$

Try: $f(0)$ and $f(3)$ to test what you found.  That is the last step.  You already did all the other steps when you found $x\le 2$ and $x=0$ as potential problems.

anonymous · Answer

wait i found x<2 not x<=2, how does this work?

e.mccormick · Answer

Ah, it is can not be less than 0.  That means what you needed to find was:

$\sqrt{1-\frac{x}{2}}\ge 0$ is a valid input.  Another way of saying that is $\sqrt{1-\frac{x}{2}}< 0$ is invalid or $\sqrt{1-\frac{x}{2}}\nless 0$

e.mccormick · Answer

Actually, I don't even need the radical.  It is finding when what is under the radical meets that.  Sorry about the confusion there.

e.mccormick · Answer

$1-\dfrac{x}{2}\ge 0$ is valid

$1-\dfrac{x}{2}< 0$ is restricted

You can think of it either way and solve.

$1-\dfrac{x}{2}\ge 0$ is valid method

$1-1-\dfrac{x}{2}\ge 0-1$ 

$-\dfrac{x}{2}\ge -1$ 

$(-2)\cdot-\dfrac{x}{2}\le (-2)\cdot-1$ 

$x\le 2$ is valid (part of the domain)

$1-\dfrac{x}{2}< 0$ is restricted method

same basic math, but I get:

$x > 2$ is restricted/invalid

If it is restricted above 2, then at 2 and below it is valid.

Whichever way makes more sense to you.

e.mccormick · Answer

Remember, when you multiply through with the -2, the < or $\ge$ changes direction to $>$ or $\le$. When you said: "for the second one i got x > 2" "oh the second one is x <2" in two messages, I misunderstood. I thought you were getting the relationship between restricted: $x > 1$ and valid $x\le 2$. Look at a graph of the original to get an idea of how this works: https://www.desmos.com/calculator/6prn6fekqy

e.mccormick · Answer

restricted $x > 2$ and valif $x\le 2$ that is...