Need math help!! Simple problem! (1 Viewer)

[Van Patton]

OK I'm a freshman in college and it's been a while since I've done this so I'm rusty! The problem is: x^2-1/2x-4. Thanks for the help!

 

[Greg_R]

Ummm... I think we need the other side to the equation! (i.e. = 0 or something).

 

[Van Patton]

Um........Well it stemmed from this. Here is the original equation.


Given that f(x)=x^2-1 and g(x)=2x-4, find each of the following, if it exists.

(f/g)(x)=

 

[Christ Reynolds]

i think it may be asking you to plug in, for every value of x in f(x), g(x). im not sure though, confused by how it is written. but if that is the case, you have (f/g)(x)=[(2x-4)^2]-1. just pretend that g(x), also 2x-4, is x in the f(x) equation, and plug it in, as i did above. you may be asking for something else entirely though

CJ

 

[Van Patton]

nah man thats good just solve that out and hoook me up with an answer. Thanks!!!

 

[Holadem]

(f/g)(x)=

This is f divided by g. There is nothing else to do other than what you already wrote, x^2-1/2x-4, unless you want to attempt long division. I don't see why though.

Also, DYOFHW. I'll let you figure that one out.

--
H

 

[Edwin-S]

Just curious, do you think you will learn to do calculus by having someone else figure out the answers for you?

 

[Bryan X]

 

[Edwin-S]

Sorry, but this is a calculus question. It is not as simple as dividing the top by the bottom. It could be a case of substituting g(x) into f(x), but it could also be question that uses f' and g'.

like f/g(x)= f(g'(x)) + f'(g(x)/g(x)^2

I'm not sure if that equation is totally correct. I don't have my calculus text available, and even though I took the course I didn't do too well. I would actually have to look it up and see if it correlates.

 

[NickSo]

The best i could do dividng F by G is reducing it by factoring the numerator...

I think its more complicated than that...

 

[Holadem]

What makes you infer such things from the simple, if perhaps erronous statement f/g(x)?

That's what I don't understand. How does someone start thinking of derivatives, with something like f/g(x)?!

--
H

 

[Morgan Jolley]

What would really be helpful is the entire text of the equation, as well as an idea of what topic this deals with.

 

[Edwin-S]

Ah, yes, I vaguely remember now. You might be correct. It could be a substitution problem too, though. Damn, I wish I had my text book and notes here. Now it is going to bother me. I agree with your other comment though.......DYOFHW.

Edit: I should add, wouldn't that require a limit to be implied? God....I can't believe I only took this stuff a little while ago and I'm already having trouble remembering how it went.

 

[Van Patton]

Its not hard.......you just divide what I gave you the first time and that is our answer. It is SO easy and all you all are just making it much more harder than it needs to be.

 

[Haggai]

The college math lecturer (me) should probably step in here. Edwin, the rule you were thinking of is the quotient rule for derivatives. It allows you find the derivative of a function that has the form f(x)/g(x), where the / stands for division (as it usually does). That rule can be expressed this way:

[f(x)/g(x)]' = [f'(x)g(x) - f(x)g'(x)]/g(x)^2

However, the original question:



does not appear to involve taking any derivatives. It's most likely just looking for f(x) divided by g(x). A similar notation can be used for other ways of combining functions:

(f+g)(x) = f(x) + g(x)
(f-g)(x) = f(x) - g(x)
(fg)(x) = f(x)g(x)

Lastly, I second (or third) Holadem's advice, DYOFHW. Ask for help, OK, but don't expect someone else to just do the problem.

 

[Van Patton]

Look I did four out of five of the same problems just like this! So far NO ONE has helped me. I could've reiterated the babble that has come out of everyones mouths so far. Please help me divide that or dont post anything.

 

[Edwin-S]

Yes. Thank you, Haggai. I finally decided to use this wonderful thing called the internet and actually look up the notation and see if I was making a complete hash of it. Sure enough I was.

I have always wanted to pass calculus and have yet to do so. I've taken the bloody 1st year course twice, but I just can't seem to get my head around it. With your explanation some of it started to come back a bit.

 

[Will Pomeroy]

If you're going to ask people to DYFHW, you could at least be polite...

Its already in its simplest form. You might want to double check the question.

 

[Edwin-S]

Yes. I factored it, and it doesn't seem to be able to be simplified beyond the original form.

 

[Haggai]

As it was written here, it would factor as:

[(x+1)(x-1)] / [2(x-2)]

No other simplifying possible from that.