Function homework help

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September 27th 2012, 01:40 PM
Kathrynm77

Function homework help

Let f be the function given by f(x)= 2x/sqrt(x^2 + x + 1)
A) find the domain and justify
B) write an equation for each horizontal asymptote
C) does the graph cross any horizontal asymototes? Show work

So I started this and got
A) domain is all reals because denominator doesn't have restrictions
B) y=2 and y=-2 however I found these graphically and need to know how to do that algebraically
C) yes it crosses y=-2 (I also found this graphically and need it algebraically)

Thanks so much in advance!!!
September 27th 2012, 01:55 PM
TheEmptySet

Re: Function homework help

Quote:

Originally Posted by Kathrynm77

Let f be the function given by f(x)= 2x/sqrt(x^2 + x + 1)
A) find the domain and justify
B) write an equation for each horizontal asymptote
C) does the graph cross any horizontal asymototes? Show work

So I started this and got
A) domain is all reals because denominator doesn't have restrictions
B) y=2 and y=-2 however I found these graphically and need to know how to do that algebraically
C) yes it crosses y=-2 (I also found this graphically and need it algebraically)

Thanks so much in advance!!!

To find the horizontal asymptotes you need to take the limits as $\lim_{x \to \pm \infty}f(x)$

First here is a hint to remember

$\sqrt{x^2}=|x| \ne x$ so be extra careful when taking the limit as $x \to -\infty$

Recall the function definition of the absolute value is

$|x|=\begin{cases} x, \text{ if } x \ge 0 \\ -x, \text{ if } x < 0\end{cases}$

Here is the positive one

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

Since we are going to postivite infinity we get that $|x|=x$ so

$\lim_{x \to \infty}f(x)=\lim_{x \to \infty}\frac{2}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} }=2$
September 27th 2012, 01:56 PM
MaxJasper

Re: Function homework help

http://mathhelpforum.com/attachment....1&d=1348783078
September 27th 2012, 02:31 PM
johnsomeone

Re: Function homework help

A) Correct. Inside the square root you have, after completing the square $((x+ 1/2)^2 + 3/4)$ which is never 0 (bad for division) or negative (bad for square root).

B) Correct. Algebraically, you evaluate the two limits: $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x)$ . Do you know how to do that?

C) This is two problems, because you have two lines (horizontal), and for each of them you're being asked if the line intersects the graph of the function.
Let's first look at the line y=-2. Suppose P is a point of intersection. What does that mean? It means that P is on the graph of the line y=-2, and P is on the graph of the function f. What does that mean? If P = (a, b), then it means that b = f(a). That's what the graph of the function y=f(x) is - points with coordinates looking like (x, f(x)) where x is in the function's domain. The same thing goes for the horizontal line y =-2. P = (a, b) is on that line means that P's y-coordinate, b, is the result of plugging P's x-coordinate, a, into the equation of the line.
OK, P = (a, b) is an intersection point of the graph of f and the graph of the horizontal line y=-2.
Therefore b = 2(a)/sqrt(a^2 + a + 1) (plugging P into the formula for the graph of f)
and b = -2 (plugging P into the formula for the graph of y = -2).
Note that when you "plug a in for x" in the formula for the horizontal line, nothing happens, because there is no x in that formula! It's the constant function that gives back a value -2 no matter which x value you give it. So when you plugged a in for x, it produced -2. Plugging b in for y and you get: b = -2. That's it.

Now solve: b = 2(a)/sqrt(a^2 + a + 1) and b = -2.

Well, you've solved for b pretty easily! Plug that into the other equation,and then solve for a.

Solve for a: -2 = 2a/sqrt(a^2 + a + 1). Do you know how to do this?

When you're done, assuming it even has a solution, you'll need to plug a back into f to see if in fact f(a) = b = -2. It might not. In other words, CHECK your answer. This isn't checking for a mistake, but for a legitimate math purpose that you'd need to do even if your were computationally infallible!

Now check the other hortizontal asymptote, y = 2. If you repeat everything above for this case, you'll end up with:

Solve for a: 2 = 2a/sqrt(a^2 + a + 1).

Notice how similar it is to the previous provblem when y=-2? They're the same except for a minus sign. To solve either equation you'll have to square it eventually (to undo that square root), and as soon as you square it you'll lose the information as to whether it started out being the first equation, or the second. You won't know whether the a you end up solving for applies to the y=2 line or the y=-2 line! The only way to tell will be to check by evaluating f(a)! This is why that check isn't optional. Squaring both sides of an equation always loses information about whether you had a + or a - prior to the squaring.