Hard Light Productions Forums
Off-Topic Discussion => General Discussion => Topic started by: Topgun on September 16, 2009, 10:13:03 pm
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look guys I reall need help, especially with math. you see, im in college. and I think I might have to drop out of math and take it again next semester. its 11:00pm here and I still have to finish these problems and START English. I pretty much do this everyday: I got to school, then I work for 2-3hrs, then I do homework until 1am to wake up at 6 and start a new day. I can't take it anymore.
so please, please help me out.
I need to find the inverse of y=3(x-2)^2 + 7[2,infinty)
so far I have (x-3)/3=Y^2 - 4y
I need to find the inverse of y=(3x-4)/(2x+5)
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inverse: x = (3y-4)/(2y+5)
:D
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I need to solve for y. I can't just write it like that.
or maybe you knew that...
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I believe the premise behind homework is that you practice and learn the material. I didn't think that in the year between my own graduation from university and now the point had become to get people on the net to do it for you.
I'm sorry to hear you are having difficulty with the workload, but if this is the route you choose to take, it will further undermine your ability to perform in subsequent courses. You'll be better served by a lighter courseload or reduced hours at work than having us post answers for you to copy.
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look guys I reall need help, especially with math. you see, im in college. and I think I might have to drop out of math and take it again next semester. its 11:00pm here and I still have to finish these problems and START English. I pretty much do this everyday: I got to school, then I work for 2-3hrs, then I do homework until 1am to wake up at 6 and start a new day. I can't take it anymore.
so please, please help me out.
I need to find the inverse of y=3(x-2)^2 + 7[2,infinty)
so far I have (x-3)/3=Y^2 - 4y
Well, I'm gonna put here a few tips but not solve the homework for you.
Zeroth, refer to the definition of mathematical function. A function is a mathematical operator that changes values from input group into output group, correct? And each input value has it's own, singular corresponding output value. Two input values may share the same output value, though.
Inverted function does the same, except it is a mathematical operator that works backwards, changing the output values of the original function into their respective input values.
However, if multiple values from input group result in the same output value, there will be problem for the inverse function since it would need to produce multiple output values and that is not possible for the concept of (single) function. This kind of functions are not directly invertible and need to be split into two - or more - parts depending on how many values of x are produced out of value of y at different ranges.
Furthermore it is possible that you end up getting results out of imaginary range depending on what you need to do during the inversion operations, which define what values of variable y are acceptable for the inverse function.
Now that you remember what to keep in mind when looking at inverse functions...
First, look at the graph of the equation (the images aren't hosted very long so if you don't see the image, go to wolfram alpha (http://www.wolframalpha.com/input/?i=y+%3D+%283%28%28x-2%29^2%29+%2B+7%29) and input the equation there to see the graph).
(http://www1.wolframalpha.com/Calculate/MSP/MSP999197a633aab922d100000192ee398ad70di5b?MSPStoreType=image/gif&s=46)
At value x=2, function f(x)=y=7. Above and below this value of x, the function gets values larger than seven for y. However, if values of y are less than seven, there simply is no corresponding x-value, so your inverse function will only be valid when y>= 7, and even then there's basically two values of x that each value of y produces, which means that for a complete solution you will need complex numbers... and depending on your level you may or may not need to include them.
Anyway... multiple values of X at same Y means that the function is not directly invertible. A function means that it produces one output value out of one input value. If it produces more, then there is a problem and it means that for an inverted function you need to divide it into two or more parts depending on how many results you get at each value of y.
As for actually solving the equation into f(y)=x form rather than f(x)=y, it is largely a simple matter of conversion - just manipulate it as long as you need to so that you have single x at one side and a function with y- variables in it on the other side.
In this case, I'll give you a hint...
Sqrt ( f(x)^2 ) = | f(x) | (bars here mean absolute value, alternative marking would be abs f(x) or +-f(x) )
What then is Sqrt ( (x - 2)^2 ) ?
I need to find the inverse of y=(3x-4)/(2x+5)
Same approach.
(http://www4a.wolframalpha.com/Calculate/MSP/MSP1238197a57eif83gg48e000017bg1hde274584d1?MSPStoreType=image/gif&s=59)
First of all you'll see that when 2x+5 = 0 due to division by zero, the function does not have valid output (or, well, lim f(x) = +- infinity depending on which side you're approaching it from) and that needs to be taken into account, of course.
Regarding the solving of the equation... I would do polynomial long division (http://en.wikipedia.org/wiki/Polynomial_long_division) first:
3/2____
2x + 5 | 3x - 4
-2x - 15/2
0x - 23/2
Therefore, 3x-4 / 2x+5 == 3/2 - 23/(2*(2x+5))...
y = 3/2 - 23 / 2(2x+5)
...and I believe you should be able to continue from that point.
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I believe the premise behind homework is that you practice and learn the material. I didn't think that in the year between my own graduation from university and now the point had become to get people on the net to do it for you.
I'm sorry to hear you are having difficulty with the workload, but if this is the route you choose to take, it will further undermine your ability to perform in subsequent courses. You'll be better served by a lighter courseload or reduced hours at work than having us post answers for you to copy.
Now that's just condescending. If there's a problem that we run into and despite hammering away at it we still aren't getting it, wouldn't the smart thing be to ask someone who knows how to do it to show you the method (or at least get you started) so you can review it? Otherwise you aren't learning, you're wasting time.
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inverse: x = (3y-4)/(2y+5)
From here it's just algebraic manipulation (no long division necessary here): Multiply by the denominator; group terms with y on one side and terms without on the other; factor out the y on the first side and divide by the other factor to get your answer.
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inverse: x = (3y-4)/(2y+5)
From here it's just algebraic manipulation (no long division necessary here): Multiply by the denominator; group terms with y on one side and terms without on the other; factor out the y on the first side and divide by the other factor to get your answer.
That is a valid method as well, and both correct ways result in the correct answer which is distinctly different from iamzack's crude substitution of x by y and vice versa. :rolleyes:
Also note that when asking mathemathics help, the answer you get might not be the perceived easiest, fastest or even simplest form. People tend to use different methods to solving problems like this, and there are usually more than one route to correct answer.
The advantage of polynomial long division is that it is sort of a brute-force solution and it's an extremely useful algorithm, so practicing it will be beneficial anyways even if you can solve the problem by simpler individual operations.
Both routes lead to the same correct answer which is
x = (-5y-4)/(2y-3)
...or, in images:
Function: f(x) = y=(3x-4)/(2x+5)
(http://img34.imageshack.us/img34/7820/78227954.gif)
Inverse function: f(y) = (-5y-4)/(2y-3)
(http://img34.imageshack.us/img34/1862/92513109.gif)
Inversion with x and y swapped with each other: f(x) = (-5x-4)/(2x-3)
(http://img3.imageshack.us/img3/6965/86920036.gif)
Note that the inverse function is not defined at value x = 3/2 while the original function is not defined at value -5/2.
Besides, the long division in this case isn't particularly long in the first place. It was the most obvious method to me.
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thanks for your help guys :thumbsup:
I don't want someone to do the homework for me, that's not the point, I just need some help to understand how to solve the problem. basically, I need a tutor.
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Asking for help is okay by me as long as you can show you done something yourself and exhausted the available options. So next time please show a little bit more of the algebraic manipulation and especially give reasons why you do each step. As long as these conditions are fulfilled I'll be glad to help.
Mika
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thanks for your help guys :thumbsup:
I don't want someone to do the homework for me, that's not the point, I just need some help to understand how to solve the problem. basically, I need a tutor.
Need a tutor, college provides.
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I use the tutors at the college whenever I can, but I live an hour away.
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It takes me an hour to get to work, just get up an hour earlier :blah:
Sorry if i sound harsh but i don't see an hour away as an excuse.
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I don't get what they are asking:
(http://i310.photobucket.com/albums/kk419/So0pahBoy/problem.png)
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Short answer: Nothing, X is the variable in that admittedly very perverse looking form of an actually very simple expression. X is what you put in it. :p
Also, there's no range given for X although quite obviously in this case X must not be zero (in real numbers that is). (Strange as this might seem, when you simplify the expression, this limitation disappears, which invokes a question - should the original limitation still be enforced...)
Anyway, the end result is what changes. What they're looking for is probably some simplification of the equation. In other words, crack that nut open.
Again, long division would be a very suitable tool here (and you will likely find it rather short example of long division in this case): P(x) = (x-1 + x-2 + x-3) / (x-4 + x-5 + x-6) and so forth... [ P(x) = polynome with x variable)
Another way is to simply look for a factor that you might be able to use to multiply x-4 + x-5 + x-6 with and end up with (x-1 + x-2 + x-3). Long division does the same algorithmically, but in some cases you can pretty easily see what factor you need for the multiplication (which will, of course, be the answer to the division as well). This is one such case, but if you can't see it, do the long division, it splits the problem up into smaller sections that are individually easier to manage.
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ah, I see now, I guess what they are looking for is x^3.
Thanks.
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I use the tutors at the college whenever I can, but I live an hour away.
Is this a community college or a full on university?
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Head spinning...can't see...makes no sense... :snipe:
Seriously, math and I don't get along. I get lost somewhere around factoring an equation and can't find my way back before I get left behind. Which results in my attendance to my math courses falling off, which results in my failure. I'm serious, I'm a Computer Science major who can't do math much beyond basic multiplication and division. The stuff just doesn't pass my logic test. Of course, it's abstract, which I am not. I need concrete values to be able to process the information presented. "X" is a letter of the alphabet and not some random meaningless character representative of a given piece of data. My brother is a math teacher, he eats this stuff up. But every time he tries to help me all it does is result in a shouting match because there are 13 different ways to do something, all of which require that you actually know more advanced procedures that what is going on in a particular equation. It's like listening to static, there's no sense in it.
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This all looks so damn familiar...
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The thing about math for me is that it's kind of like exercise (exactly like exercise, but for the brain, in fact), and there's a question of breaking the pain barrier. As I said in IRC once, the moment a load of letters and polynomials and greek letters appear on a page my brain tends to wave a little white flag without even trying.
I think the trick is to not feel over-powered by the fact it looks complex at first glance, and, instead, try to break it down into small parts, which is a perfect attitude to take if you are a programmer anyway. I think it's a bad idea to look at an equation and expect some sort of answer to leap out at you, it's not like English or General Knowledge, where you either know the answer or you don't, math requires a process in order to resolve the question, and, especially when you get into your 30's like me, you tend to rely too heavily on experience, and not enough on actually working through the problem.
Just like exercise, 50 push ups is hard to do, but if you do 50 push-ups a day for a few months, it becomes a lot easier, as your body becomes used to the work involved. the same thing applies to your brain, exercise it regularly and it becomes better at dealing with the problems it faces.
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I can't wait u ntil you get to derivatives :drevil:. Derivatives are FUN!
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Argh, I've had enough of derivatives and integrals this year. :ick:
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Derivatives are fun. Integrals, on the other hand, are torture.
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/shrill-hoarse-girly-voice
Flee! FLEE!
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Derivatives are fun. Integrals, on the other hand, are torture.
I'm tired of both...
At least integrals for the most part are over for me.
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Im trying not to get too much help from you guys, but I can't get this for the life of me.
oops, forgot something important.
sin (t) = a, cos (t) = b, tan (t) = c
7sin(-t) - sin (t) =?
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If I recall correctly, the graph of sin(x) is rotationally symmetric if you rotate 180 degrees around the origin, so shouldn't sin (-t) = -sin(t), making it -7sin(t) - sin (t) = -8 sin (t)?
That's all assuming sin (-t) = - sin (t). You should know that by now, if not look it up in your textbook.
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eh, ah... your right, I was thinking of cos. cos doesn't care if its positive or negative.
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Math is awesome.
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I have honors trig this year. As in, it would be honors if i was in 11th grade :ick:
I may wind up posting here.
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SEriously, i hate to put my foot down, but HLP is not a "WAGHHH WAGHHHHHHHH, MATH IS HARD WAGHHHH" forum, All due respect etc lalalal. but what's stopping you from getting the tutorage that you get from your current institution of learning?
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Me? No free periods, not enough time after school. I meant if I was having trouble understanding a concept (my textbooks suck), I might ask here.
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Not you Titan, you're a well meaning fine upstanding yong innocent,,,,,,,
I just read this whole thread as a learning is lazy,m do my stuff for me thing.....
from post
1.........
ps guinness woins :p