I have a problem that says, as x goes toward infinity, the function approaches (?) The rational function is f(x)= 4x^2-1/4x^2-9 How do I figure out how to answer that!? Help please! =(
then you have juste to calculate 4x^2/4x^2 =1
Here's how to do it... \[\Large \frac{4x^2-1}{4x^2 - 9}\]
can you tell me why? I don't really get it if i don't know why.
Figure out the largest exponent of x, whether in the numerator or the denominator. .. so... what IS the largest exponent of x? ^_^
2
That's right. So what I want you to do now, is multiply that entire thingy with this *ugly* expression... \[\Huge \frac{\frac1{x^2}}{\frac1{x^2}}\]
By the way, that *ugly* expression up there^ is effectively equal to 1
so just multiply it by 1?
Yeah... but a rather special incarnation of 1. You'll see why later...
You can also do it by just ignoring everything but the highest powers of x on the numerator and denominator (these become the dominant terms as x gets larger) in this case it's... \[\Large \frac{ 4x^2 }{ 4x^2 }= \]
\[\huge \frac{4x^2 - 1}{4x^2 - 9}\cdot \frac{\frac1{x^2}}{\frac1{x^2}}\] And this is just equal to \[\Huge \frac{\frac{4x^2}{x^2}-\frac1{x^2}}{\frac{4x^2}{x^2}-\frac9{x^2}}\] Which just looks horrible doesn't it? :P
i'm not very good at multiplying fractions and things. i'm pathetic really. failing college algebra for the 6th time.
But do you see what I did there? I just multiplied everything (in the numerator and denominator) by \(\Huge \frac1{x^2}\)
Then do my method @kawriter
Yeah, I suppose ^ I do love the magic, though :3
i'm not sure I see how you got what you got.
is that the only step @agent0smith ?
Look at the highest powers of x on the top and bottom. Ignore everything else.
You have to simplify what i gave.
okay
The thing is, @kawriter as x gets larger and larger, those 'lesser' exponents become more and more insignificant, and that the value of the expression depends more on the x with the highest exponent...so you really only have to check how that 'highest-exponent' term behaves ^_^
@agent0smith doI simplify it by square rooting the whole thing and coming out with 2/2 = 1?
crap! now i have to graph that function... If i'm asking too much of anyone, just let me know!
Yes. Another example: \[\Large y = \frac{ -2x^3 + 3x^2-x+1 }{ 4x^4-7x^3-5 }\] First identify the highest power on the top and bottom.
3
and 4
Well what are the terms with the highest powers? Graph: https://www.google.com/search?q=(4x%5E2-1)%2F(4x%5E2-9)&oq=(4x%5E2-1)%2F(4x%5E2-9)&aqs=chrome.0.69i57j69i62l3.265j0&sourceid=chrome&ie=UTF-8
\[\Large y = \frac{ -2x^3 + 3x^2-x+1 }{ 4x^4-7x^3-5 }\] highest power of x on the top is the -2x^3. What about the bottom?
4^4
that graphing page is AMAZING!! THANK YOU!
You mean 4x^4?
yes sorry.
Now divide them like i did above, for the first question.
so would that be -1/2x?
Yep. And what is -1/2x close to if x is huge?
um, i have no idea.....3?
When you divide a small number by a huge number, what's the result close to?
the huge number
okay, i think.
So a small number divided by a huge number is close to...?
an even smaller number?
Yes, but you could round it to what?
1?
Well is a number like 0.000000001 closer to 0 or 1?
oh, 0
Which means -1/2x, if x is huge, is basically equal to...
0
Yep. What about 10/2x^2, if x is huge, it must be basically...
0
so for my initial problem, the function approaches 0?
You worked out the answer to the original one earlier... "doI simplify it by square rooting the whole thing and coming out with 2/2 = 1?"
It's only equal to zero if AFTER simplifying, you end up with an x on the bottom, like -10/3x.
gotcha. i just did another one and I got 10/x so that equals 0?
Yes. And if you end up with something like x/5... what result do you get if you divide a huge number by a small number?
a really big number? cause x isn't on the bottom.
Yes, which just means the function approaches infinity as x approaches infinity.
coooooollllll!!! makes sense!
and the one i did earlier,
i had 4x^2/4x^2
so i took the square root of both, x^2's cancelled out and I was left with 2/2 which =1
You can't take the square root of anything, all you can do is cancel off common factors. The 4s cancel, and the x^2 both cancel, leaving you with 1.
ohhhhhhhh.....gotcha. okay. why isn't it 0 if both cancel?
Well... is 4 divided by 4 equal to zero?
oh. you make sense sensei. lol. i'm super dumb in math! sorry!
Remember that x is just a number. So is x^2. 20x^5/20x^5 = 1
ahhh....cool.
thanks @agent0smith you are AWESOME!
You're welcome :)
can i ask you a new question?
Sure.
is the remainder theorem like synthetic division?
if it's not i'm so screwed. ha.
Well, you can use synthetic division to get a result/remainder, if that's what you mean.
okay, i think that's what i mean. heh. i'm given an equation and i have to use the theorem to find f(-3)
Yeah, you can use synth division. The remainder will be the value of f(-3)
oh okay cool.
starting synth division always messes me up. do you start out multiplying?
Like this: http://www.purplemath.com/modules/polysolv/remndr02.gif The first coefficient, you just bring it down, then multiply by the number you're dividing by.
would you mind checking it?
i'm so bookmarking that page! thanks
It has several examples of synthetic division at the bottom. What was the actual question?
Given 9x^4 + 12x^3 +2x^2 -7x +74, find f(-3)
Did you get 518? http://www.wolframalpha.com/input/?i=+f%28x%29+%3D+9x%5E4+%2B+12x%5E3+%2B2x%5E2+-7x+%2B74%2C+find+f%28-3%29
no, I got 292 =(
Ahhh!!! I don't know what I did wrong! =(
You could take a pic of your work and upload it on photobucket or something, and paste the link here so i can check.
my phone can't take pictures. =(
Can I maybe describe my work?
i had the -15 as -39. wrong from the start. okay i'm going to work it one more time without looking at the picture.
Fixed an error |dw:1374754519644:dw|
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