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[Nostalgia]

Thank you very much. I did realize what I was doing wrong before the test, but seeing this helps to confirm that I at least got it right by then, so I appreciate it. XD

 

[Dogenzaka]

I kind of suck at math. Could someone walk me through this problem?

Show all of the zeroes that lie between [-3, 3] for
f(x) = 2x^5 - 13x^3 + 2x - 5

Do I just plug in -3 and 3 into the (x) and solve or what...?

I'm really bad at functions...does anyone have a good place for advice on functions?
Thanks.

 

[Banishing Blade]

The idea is to set f(x) = 0, and then solve for x. There's gotta be an easier way than brute-force factoring this, though. What's the current topic in your class (or rather, the topic that you got this question from)?

 

[Teiku 5]

LAZY WAY
~~~~~~
take your calculator.
put the equation in Y=.
hit 2nd -> Trace (which is Calc) and hit the "zero" option.
choose left bound a little left of every y=0 on the graph and right bound a little right of every 0. do this until you have them all.

i forget the real way for a problem like that. let me think a minute.

 

[Nyangoro]

EDIT: Scratch that last statement (I forgot to count in fractions, silly me). You want to find the rational zeroes.

To find the possible rational zeros, you take the multiples of constant number (in this case 5) and divide them by the multiples of the leading coefficient (in this case 2)

Which would be +-1,+-5 divided by +-1,+-2

So, we've got:

+-1
+-1/2
+-5
+-5/2

*You always include both positive and negative options.

Since you are looking for all values of x between [-3,3] that make f(x) = 0, you can eliminate +-5

Plug in the other six and whichever one(s) equal zero is (are) your answer(s)

 

[Banishing Blade]

EDIT: Scratch that last statement (I forgot to count in fractions, silly me). You want to find the rational zeroes.

To find the possible rational zeros, you take the multiples of constant number (in this case 5) and divide them by the multiples of the leading coefficient (in this case 2)

Which would be +-1,+-5 divided by +-1,+-2

So, we've got:

+-1
+-1/2
+-5
+-5/2

*You always include both positive and negative options.

Since you are looking for all values of x between [-3,3] that make f(x) = 0, you can eliminate +-5

Plug in the other six and whichever one(s) equal zero is (are) your answer(s)

Problem is that none of those values result in f(x) = 0.

 

[Nyangoro]

Hmm . . . at least one of them should (assuming I have the correct idea). I'd say that they might be all imaginary, but that's too convenient for the problem, lol.

Dogen, you copied the formula correctly, right?

 

[Dogenzaka]

According to the paper, yeah.

To find the possible rational zeros, you take the multiples of constant number (in this case 5) and divide them by the multiples of the leading coefficient (in this case 2)

Which would be +-1,+-5 divided by +-1,+-2

Err....so you're saying if I ever want to find the "rational zero" of a function, I do that?

Well why did you choose only one of those terms and how did you choose that term to start your work with?

I'm assuming you chose 2x^5...but why that one? And why would that one lead me to rational zeros?

BTW maybe the problem is wrong, I dunno, but that's what I copied lol.

 

[D∆NTE]

Is'nt this considered as a form of cheating.

 

[Nyangoro]

According to the paper, yeah.



Err....so you're saying if I ever want to find the "rational zero" of a function, I do that?

Well why did you choose only one of those terms and how did you choose that term to start your work with?

I'm assuming you chose 2x^5...but why that one? And why would that one lead me to rational zeros?

BTW maybe the problem is wrong, I dunno, but that's what I copied lol.

Well, I know it's a process for finding zeros, but I could be off somewhere, lol.

Anyway, you take the multiples of the Constant (the number that doesn't have a letter attached to it) and divide each of them by each multiple of the Leading Coefficient (the number attached to the letter with the higher power). If it helps to visualize it like this, take the multiples of the number on the far right end and divide them by the multiples of the number on the far left end. So, let's say the Constant is 4 and the Leading Coefficient is 2.

Multiples of 4 = 1,2,4 (because 1 x 4 = 4; and 2 x 2 = 4, and there's no other way to multiply them and get four)

Multiples of 2 = 1,2

So you'd take 1/1 , 2/1 , 4/1 , 1/2 , 2/2, 4/2

And because you take both the positive and negative values, the possible rational zeros are:

+-1/2, +-1, +-2, +-4 (since 1/1 & 2/2 are the same, and 2/1 & 4/2 are the same, you only need to count them once)

The process should give you numbers that make f(x) = 0 , unless all possibilities are imaginary numbers. It seems odd that you'd get a problem with all imaginary solutions, but I guess it's not impossible (really unlikely though). Just make sure you plug in the same number both times for the positive and negative values.

 

[Dogenzaka]

Is'nt this considered as a form of cheating.

Not if it's a practice problem set, and the only assignments that count for weighted average grades are actual exams >_>

And what if there are no terms in the function that are constants? What if every term has a variable such as x?

So you're saying, with that problem, you picked the 2 from 2x^5 because it's the Leading Coefficient (has the highest "power" of all the terms) and you picked the 5 from -5 because it's a Constant?

 

[Nyangoro]

Yeah; and if the last number ends in with a variable attached to it, you can factor out an x (or whatever letter it is) from the problem.

 

[krexia]

Just for reference, I used an online polynomial calculator and got x = -2.485 or -0.831. So solutions exist, and they are between -3 and 3. But they're not nice numbers and I don't recognise those decimals. It may be that brute factorisation is the only way to solve this.

 

[Dogenzaka]

The problem is I hardly know factorization ;_; does anyone know a good source to learn about it?

 

[▽]

Damn, i don't think you are able to solve it by hand, i've heard my math lecturer talk how it's becoming nigh impossible from fourth order equations and higher. Hell, third order equations need 1 to be a solution to solve them, since that would leave you with a quadratic equation. If you were able to use substitution, it would be possible, but that's impossible with these terms. I'd stop bothering with this one and go on with equations which are at least reasonably possible.

 

[krexia]

Yeah, I've just had a play around with the equation and can't find any straightforward way to do it by hand. Like ikkuh says, even solving third order polynomials involves figuring out one of the factors. Since the solutions here are horrible numbers, there's no nice factor to guess.

I reckon one of the coefficients or signs in the equation is wrong on the paper. If you want help on less horrible functions I'd be happy to talk you through how factorisation usually works

 

[Teiku 5]

usually i don't have any problems with math but.....

we're finding limits using infinity and this is just an extension it.

i need to find the horizontal and vertical asymptotes for the graph of f (algebraically, not by looking at a graph):

f (x) = ln(1 + e^x)

i can't do the distributive property because it's a logarithm so that's not even what it means. i think i can find the vertical ones by setting it equal to zero, but i still don't know how to go bout finding the horizontal one.

EDIT: figured it out. there was no vertical asymptote which makes sense. the horizontal one is the one you find by setting it to zero.

 

[zachen]

does anyone know of important goverment buildings in india please post a pic and a link

 

[Nyangoro]

Well, there's the Bombay High Court

Bombay High Court Official Website

I love Google. So fast, so easy.

 

[zachen]

Well, there's the Bombay High Court

Bombay High Court Official Website

I love Google. So fast, so easy.

Thank you now all i have to do is type up a page and 1/2 report than i can go play infamous woot^.^