In what interval is the function f(x) = (square root)(x^2+5x+4(end square root)

Question

debbieg · Answer

in what interval is $\large f(x)=\sqrt{x^2+5x+4}$...... what?

anonymous · Answer

what's the interval notation that it's in

anonymous · Answer

an explanation would be nice too :)

debbieg · Answer

I don't know what you mean, "in what interval". Usually we would ask something like:

"in what interval is f(x) increasing?" ... or... "in what interval is f(x) decreasing?"

Or maybe you mean the domain of f(x)??

I don't know what the question means, "in what interval is f(x)."

debbieg · Answer

And I only give explanations, not answers. :) but I can't explain without knowing what the question is asking.

Have you given the EXACT question, as it is asked?

anonymous · Answer

i mean what's the interval notation in regards to domain, and that is the exact question, i was confused at first too

debbieg · Answer

Wow that goofy wording to be asking for domain, lol... but ok, THAT we can work with.

OK, so you have some stuff under a square root, right? So in terms of domain, what do you have to worry about? What condition has to be true?

anonymous · Answer

i have no clue...

debbieg · Answer

If $\large f(x)=\sqrt{x}$, what is the domain? What must be true about x?

anonymous · Answer

[0,infinity)

debbieg · Answer

OK, right... must have $x \ge0$. good.

so here, we have the function:

$\large f(x)=\sqrt{x^2+5x+4}$

so the stuff under the square root HAS to be non-negative, that's what you need to make sure of. In other words, when you figure out for what x's it is true that:

$\large x^2+5x+4 \ge0$

..then you'll have your domain. Follow so far?

anonymous · Answer

I didn't think about it that way, thanks so much

debbieg · Answer

So the x-values where $\large x^2+5x+4 \ge0$ is exactly what your domain is.

Do you have any thoughts on how to find where that is true? How do you solve an inequality like that, for a quadratic?

debbieg · Answer

ok, are you good now?

anonymous · Answer

$$\frac{ x-7 }{ x ^{2}-1 }$$

anonymous · Answer

Can you help me determine the domain and range of that one? It also asks for horizontal and vertical asymptotes

debbieg · Answer

Sure.... domain first.

It's a rational function, so what do we have to worry about on THIS one? (just like for a square root, we have to worry about one particular thing, when we have a rational function, there is one specific condition that we need to worry about when it comes to domain...)

anonymous · Answer

The denominator cannot be 0, or there'd be a divide by zero error?

debbieg · Answer

Exactly. So, find where the den'r=0, and you know what's NOT in your domain.

That's the only domain "issue" with this kind of function, so everything else IS in the domain. In other words, the domain is all real numbers EXCEPT those that make the den'r=0.

debbieg · Answer

So you just need to solve $x^2-1=0$

The solutions to that are NOT allowed in the domain.

anonymous · Answer

So 1 & -1?

debbieg · Answer

Exactly! which leads us to.... the vertical asymptotes. Because those occur exactly at the x-values that make the den'r = 0.

anonymous · Answer

x=1, x=-1 for vertical asymptotes?

debbieg · Answer

Yup, exactly.

debbieg · Answer

For horizontal asymptotes, there are some easy rules that depend on the degree of the num'r vs. the degree of the den'r.

Here, the degr of the num < degr of the den'r.... in that case,  you will always get a HA at y=0 (the x-axis).

Think about it, and it will make sense: as x gets BIG, the den'r will "take over" the num'r, because den'r has higher degree. So it bullies the num'r out of the way :) and the ratio goes to 0.

anonymous · Answer

is that the only horizontal asymptote?

debbieg · Answer

There is NEVER more than one HA. :) (And there might not be any.)

Why? Because the HA is the END behavior of the function.... what y-value the function is "settling in on" for very large x.  Well, at some point, it has to "settle"! E.g., it can't go to two DIFFERENT y-values, right? (If you understand what makes a relation a function, then you know why - if it settled at two y-values, then it wouldn't be a function).

anonymous · Answer

yeah, i get it, thanks so much

debbieg · Answer

You're welcome, happy to help. :)

goformit100 · Answer

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