Could someone please explain how to find the domain and range of: 1/square root (x^2+4). Without graphing. Thank

The domain is all reals; x can take any real value.

How do you work this out? Well, you start by asking yourself "are there any points or regions at which the function is undefined?" Since the function is a rational function (i.e. a fraction), an obvious place to start is by asking yourself for what values of x will the denominator be zero (and cause a problem)?

But a quick glance at the denominator should tell you that the denominator can never be zero for any value of x; specifically, the denominator is always 2 or higher.

Now, for any other possible points at which the function becomes undefined, note that there is a square root. Since we're considering only real numbers, recall that we cannot find the square root of a negative number. So you should now ask yourself for what values of x is x^2 + 4 < 0 ? The answer is none. So the square root doesn't cause any problems in the function.

So there are no values of x such that the function will become undefined, and therefore the domain of the function is all reals.

Hope this helps.

Oh ok I get it thanks so much. I’m not sure how to find the range though.

The Range would be

$0

You should think about what the lowest value you could have in your denominator, in that case it would be $\sqrt{4} = 2$

That lowest value would give you the highest value in the function (the maximum)

Now think about the largest possible value in the denominator, that would be infinity since x could always increase without bound.
This means the lowest value of the function is 0. (Now it actually doesn't get to the value 0, this way the range doesn't include that value)

The reason it can't be negative is that when you take a square root, you can only take a square root of a positive number, which only gives a positive value.

Ok thanks I understand how that works now, but then how would I find the range of 1/ squareroot(x^2+5x+6)? because there’s a 5x too

Originally Posted by happy10
Ok thanks I understand how that works now, but then how would I find the range of 1/ squareroot(x^2+5x+6)? because there’s a 5x too
$y>0$

Since

$y \rightarrow \infty \\ \\ \ \ as \ \ x \rightarrow -2 \ \ or \ \ x \rightarrow -3$

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