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[post-S]

Any help with this problem:

f(x) = (x-1)^(-1)
f'(x) = -(x-1)^(-2)

Then it's just substitution

Edit: ah limit definition
So use lim(h->0) ((x+h-1)^(-1)-(x-1)^(-1))/h

 

[Qurupeke]

Any help with this problem:

You have to use the limit of h->0 for (f(xo+h)-f(xo))/h. For example, for xo=-2 you'll end up with the limit of h->0 for 1/[3(h-3)]= -1/9, which is f'(-2).

 

[Majestad]

You have to use the limit of h->0 for (f(xo+h)-f(xo))/h. For example, for xo=-2 you'll end up with the limit of h->0 for 1/[3(h-3)]= -1/9, which is f'(-2).

If you don't mind, can you explain this a little more. I'm getting confused with the symbols.

xo==

 

[Qurupeke]

If you don't mind, can you explain this a little more. I'm getting confused with the symbols.

xo==

It's not really a xo but more like x0, sorry for that. In any case, it's any number. In your case, you want to do that 4 times, for xo=-2, xo=-1, xo=2, xo=5. The first one is lim h->0 [f(h-2) -f(-2)]/h, which should be easy to find. You do the rest like this.

 

[Majestad]

It's not really a xo but more like x0, sorry for that. In any case, it's any number. In your case, you want to do that 4 times, for xo=-2, xo=-1, xo=2, xo=5. The first one is lim h->0 [f(h-2) -f(-2)]/h, which should be easy to find. You do the rest like this.

Ah haha, thanks man, I got the answers for the rest. I know I'm a pain, but can you help me with this related problem? (that's it I promise )

 

[Qurupeke]

Ah haha, thanks man, I got the answers for the rest. I know I'm a pain, but can you help me with this related problem? (that's it I promise )

You do the same as the previous ones for xo=10 but you don't use the limit and instead you'll find something that has h in it and it's exactly like the simplified form with a and b, but with numbers in their place. Then just see what number is a and what number is b and you're done.

 

[ohlawd]

No, you were supposed to replace x in f(x) with the entire equation for g(x). Then simplify.

For example, f(x) = 3x^2 + 7, g(x) = (x+2)^2 + 1

f(g) = 3[(y+2)^2 + 1]^2 + 7
= 3 y^4+24 y^3+78 y^2+120 y+82

welp, guessed I fucked up hardcore

if it was a simple f(g(x)) then yeah I would have put in g(x) into f. that y in there threw me off. didn't know what to do exactly

all I had to do was switch x for y? shit lol

 

[cpp_is_king]

welp, guessed I fucked up hardcore

if it was a simple f(g(x)) then yeah I would have put in g(x) into f. that y in there threw me off. didn't know what to do exactly

all I had to do was switch x for y? shit lol

Yep, f(x) and f are as equal as 5 and 5. The only difference is what letter you write. For nested functions work from the inside out. If f(x) = x^2 then f = y^2. From g(f) = g(y^2) by replacing f on the inside with y^2, which just means plug y^2 in for all the variables in g

 

[Kieli]

So apparently imaginary roots come in groups of 2. I did not know that.

 

[cpp_is_king]

So apparently imaginary roots come in groups of 2. I did not know that.

Makes sense if you think about it. In order to make an imaginary number real you need to multiply it by another imaginary number. So your polynomial couldn't possibly have real coefficients unless each imaginary number had a counterpart to make it real

 

[Kieli]

Makes sense if you think about it. In order to make an imaginary number real you need to multiply it by another imaginary number. So your polynomial couldn't possibly have real coefficients unless each imaginary number had a counterpart to make it real

Mind blown.... I feel like this is something I should have known years ago, lol.

 

[cpp_is_king]

Mind blown.... I feel like this is something I should have known years ago, lol.

Corollary: If the degree of the polynomial is odd, then ______________.

 

[kgtrep]

Corollary: If the degree of the polynomial is odd, then ______________.

pray to math god that it is 1 or 3 cause Galois says we are doomed otherwise.

 

[LordAmused]

Hello MathGaf!

I am taking my first undergrad calculus course and boy, it is tough! I have never been great at maths but I've heard that hard work can really pay off.

I have a few questions though, here is one of them.

How do I find the difference quotient on no.28? The fraction really puts me off, I don't know what to do with it.

 

[cpp_is_king]

Hello MathGaf!

I am taking my first undergrad calculus course and boy, it is tough! I have never been great at maths but I've heard that hard work can really pay off.

I have a few questions though, here is one of them.

How do I find the difference quotient on no.28? The fraction really puts me off, I don't know what to do with it.

There's nothing special about the fraction. What is f(x)? It is (x+3)/(x+1). What is f(1)? It is (1+3)/(1+1) = 2. Therefore (f(x) - f(1)) / (x - 1) = [(x+3)/(x+1) - 2] / (x + 1)] = [(x+3)/(x+1) - 2(x+1)/(x+1)] / (x+1) = [(x + 3 + 2x + 2)/(x+1)]/(x+1) = (3x+5)/(x+1)^2

 

[kgtrep]

There's nothing special about the fraction. What is f(x)? It is (x+3)/(x+1). What is f(1)? It is (1+3)/(1+1) = 2. Therefore (f(x) - f(1)) / (x - 1) = [(x+3)/(x+1) - 2] / (x + 1)] = [(x+3)/(x+1) - 2(x+1)/(x+1)] / (x+1) = [(x + 3 + 2x + 2)/(x+1)]/(x+1) = (3x+5)/(x+1)^2

Don't forget to distribute the minus sign. We get (-x + 1) / ((x + 1)(x - 1)) = -1/(x + 1).

To OP, as you may know, you can interpret that difference quotient as the slope between the points (1, f(1)) and (x, f(x)). With calculus, it often helps to draw pictures and give a geometric interpretation.

 

[cpp_is_king]

Don't forget to distribute the minus sign. We get (-x + 1) / ((x + 1)(x - 1)) = -1/(x + 1).

To OP, as you may know, you can interpret that difference quotient as the slope between the points (1, f(1)) and (x, f(x)). With calculus, it often helps to draw pictures and give a geometric interpretation.

Yea thanks for the correction. I was trying to paste that expression into Wolfram Alpha so I could screenshot it written out more cleanly, but it's down for me or something right now.

 

[spelltropy]

Struggling with this question: trying to find an example of countably many open sets in (R, |x|) such that their intersection is not open.

Any ideas?

 

[KharmaPolice]

Struggling with this question: trying to find an example of countably many open sets in (R, |x|) such that their intersection is not open.

Any ideas?

I'm not sure what you mean by (R, |x|). Could you please give a definition?

 

[kgtrep]

I'm not sure what you mean by (R, |x|). Could you please give a definition?

I believe the notation refers to which space and which metric we are considering to answer the statement.

I think the classic example is the intersection of the sets (-1/n, 1/n), where n is a natural number (hence, countable).

We know a set containing one element (a singleton) is not an open set in R with the usual metric. So the question boils down to coming up with a bunch of open sets such that their intersection will be a singleton.

 

[kgtrep]

I decided to make a math blog, turn things I learned and found interesting into writing. I suppose it's like a resolution, in that who knows how long I will write on a regular basis (biweekly sounds good).

If you guys are interested in reading, just PM me and I will be happy to send you a link. Cheers.

 

[Stumpokapow]

I decided to make a math blog, turn things I learned and found interesting into writing. I suppose it's like a resolution, in that who knows how long I will write on a regular basis (biweekly sounds good).

If you guys are interested in reading, just PM me and I will be happy to send you a link. Cheers.

You can post it in here--don't worry about the self-promotion issue

 

[Dereck]

Why can't a graph be symmetrical to the x-axis?

 

[kgtrep]

You can post it in here--don't worry about the self-promotion issue

Haha, all right, if you say so.

crunchingnumbers.live

Why can't a graph be symmetrical to the x-axis?

It can, but the graph wouldn't be of a function that is a one-to-one function.

 

[Therion]

Why can't a graph be symmetrical to the x-axis?

The graph of (almost) any function f(x) can't be symmetric with respect to the x-axis, because that would mean every point (x,y) on the graph has a matching point (x,-y). So f(x) would equal both y and -y, violating the definition of a function of x.

Of course, the real answer is that the graph of a function actually can be symmetric with respect to the x-axis, provided that function is f(x) = 0.

 

[LordAmused]

Don't forget to distribute the minus sign. We get (-x + 1) / ((x + 1)(x - 1)) = -1/(x + 1).

To OP, as you may know, you can interpret that difference quotient as the slope between the points (1, f(1)) and (x, f(x)). With calculus, it often helps to draw pictures and give a geometric interpretation.

Thanks a lot for the answers gentlemen.

 

[cpp_is_king]

Haha, all right, if you say so.

crunchingnumbers.live



It can, but the graph wouldn't be of a function that is a one-to-one function.

In fact, the graph wouldn't even be of a function at all, since a function can only map a value of x to a single value of y.

The exception to this rule is the function f(x) = 0 over any subset of the reals.

 

[kgtrep]

In fact, the graph wouldn't even be of a function at all, since a function can only map a value of x to a single value of y.

The exception to this rule is the function f(x) = 0 over any subset of the reals.

True that, but I think it depends on how we look at a function. I see it as an input-output device, which may allow multiple outputs for an input, and the graph of a function is just a visualization of the input-output pairs.

The useful ones are those that give one output for each input. But multi-valued functions are considered too, at least in fracture mechanics. Think about tearing a piece of paper through the middle and pulling the two "legs" in different directions. It'd be nice to describe that the points that were initially in the middle have moved to both legs, which creates a new line (surface in 3D). We can do so with multi-valued functions. (Then we limit the inputs so that the function is 1-1.)

 

[cpp_is_king]

True that, but I think it depends on how we look at a function. I see it as an input-output device, which may allow multiple outputs for an input, and the graph of a function is just a visualization of the input-output pairs.

The useful ones are those that give one output for each input. But multi-valued functions are considered too, at least in fracture mechanics. Think about tearing a piece of paper through the middle and pulling the two "legs" in different directions. It'd be nice to describe that the points that were initially in the middle have moved to one leg or the other. We can do so with multi-valued functions. (We then limit the inputs so that the function is 1-1.)

I was just being pedantic i mean sure you can think of functions as mappings, but the formal definition of "function" demands that it be single valued. Ironically the term for a multi valued mapping is "multifunction", even though if you adhere strictly to definitions a multifunction is not a function, lol.

Anyway I imagine the person asking about graphs doesn't care about all this and for good reason - because it kind of doesn't matter unless you're trying to publish a paper

 

[kgtrep]

I was just being pedantic i mean sure you can think of functions as mappings, but the formal definition of "function" demands that it be single valued. Ironically the term for a multi valued mapping is "multifunction", even though if you adhere strictly to definitions a multifunction is not a function, lol.

Anyway I imagine the person asking about graphs doesn't care about all this and for good reason - because it kind of doesn't matter unless you're trying to publish a paper

Those damn academics.

 

[Daria]

I feel silly for bashing my head over this review question.

A course average between 75 and 79 will give result in a C. Four exams (90, 82, 70 and 78) with the last final having twice the weight. what range will result in a C+?

I set it up the following way:

1. [B]75 ≤ 90 + 82 + 70 + 78 + 78 / 5 ≥ 79[/B]. : total exam scores divided by amount of exams

2.[B] 75 ≤ 398 / 5 ≥ 79[/B]. : multiply everything by 5, correct? 

3. [B]375 ≤ 398  ≥ 395[/B] : then I'm lost

Step 3 is where I seem to get confused because normally I had x in for the fourth score. Even when I do the full division of 5 for the total exam score, I get

[B]375 ≤ 79.6  ≥ 395 [/B]in which I would subtract [B]79.6[/B] from both. 

Resulting in [B]295.4 ≤ 0  ≥ 315.4[/B] which seems to be very wrong.

 

[Omega Ultimus]

I feel silly for bashing my head over this review question.

A course average between 75 and 79 will give result in a C. Four exams (90, 82, 70 and 78) with the last final having twice the weight. what range will result in a C+?

I set it up the following way:

1. [B]75 ≤ 90 + 82 + 70 + 78 + 78 / 5 ≥ 79[/B]. : total exam scores divided by amount of exams

2.[B] 75 ≤ 398 / 5 ≥ 79[/B]. : multiply everything by 5, correct? 

3. [B]375 ≤ 398  ≥ 395[/B] : then I'm lost

Step 3 is where I seem to get confused because normally I had x in for the fourth score. Even when I do the full division of 5 for the total exam score, I get

[B]375 ≤ 79.6  ≥ 395 [/B]in which I would subtract [B]79.6[/B] from both. 

Resulting in [B]295.4 ≤ 0  ≥ 315.4[/B] which seems to be very wrong.

Sorry if you already tried this or if it makes no sense (I may be a bit rusty with inequalities), but can't you just divide 398/5 in the second step, giving you:

75 ≤ 79.6 ≥ 79 ?

 

[Montresor]

I feel silly for bashing my head over this review question.

A course average between 75 and 79 will give result in a C. Four exams (90, 82, 70 and 78) with the last final having twice the weight. what range will result in a C+?

I set it up the following way:

1. [B]75 ≤ 90 + 82 + 70 + 78 + 78 / 5 ≥ 79[/B]. : total exam scores divided by amount of exams

2.[B] 75 ≤ 398 / 5 ≥ 79[/B]. : multiply everything by 5, correct? 

3. [B]375 ≤ 398  ≥ 395[/B] : then I'm lost

Step 3 is where I seem to get confused because normally I had x in for the fourth score. Even when I do the full division of 5 for the total exam score, I get

[B]375 ≤ 79.6  ≥ 395 [/B]in which I would subtract [B]79.6[/B] from both. 

Resulting in [B]295.4 ≤ 0  ≥ 315.4[/B] which seems to be very wrong.

Daria, why are you assuming the score with 78 is the one that's double-weighted?

The way I'm interpreting the question, it looks like there are four known exam scores: 90, 82, 70 and 78. Then theyre saying there's a fifth unknown exam score whose weight is double.

Also, I think the back-end of your statements are incorrect. I.E you are writing "≥ 79" at the end of your first statement. That seems backwards. Should be "≤ 79" to make logical sense.

Lastly, they mention an average of between 75 and 79 is a C... specifically because they use the word "between" I wouldn't use "greater than or equal" signs or "lesser than or equal" signs in your opening statements. Just regular "greater than" and "lesser than" signs.

So I would do:

75 < (90 + 82 + 70 + 78 + 2x) / 6 < 79

Where x represents the unknown fifth exam score. The above statement says "My 5 exams when averaged out will need to score between 75 and 79 to get a C grade."

Multiply all sides by 6.

460 < 320 + 2x < 484

Subtract 320 from all sides.

140 < 2x < 164

Divide all sides by 2.

70 < x < 82

So now we know minimum final exam score is "greater than 70" for an overall C average. And maximum final exam score is "less than 82" for an overall C average. We want a C+ as the overall average so we want the opposite of "less than 82".

So I'd say my final answer is x &#8805; 82 to guarantee a C+ minimum. The question actually asks for the range of grades that give you a C+ but that seems impossible to determine - the grade at which a C+ becomes a B- could be arbitrary.

 

[Two Words]

You and 5 other people are going to play Russian Roullete. You choose to go last. Persons 1 through 4 all survive their turn. Person 5 asks you if you want to switch turns. Do you say yes or no? Does your answer effect your probability for survival?

 

[kgtrep]

You and 5 other people are going to play Russian Roullete. You choose to go last. Persons 1 through 4 all survive their turn. Person 5 asks you if you want to switch turns. Do you say yes or no? Does your answer effect your probability for survival?

I'd say no. I will be 5 times more likely to die by switching turns. (I may be wrong, so please correct me if so.)


----


In the Monty Hall game, the host presents you 3 doors, behind one of which contains a prize. Once you pick a door (say door #1), the host opens one of the other two doors to show that it's empty (say door #3). The host then gives you a choice: do you stay with door #1 or switch to door #2?

Had we no knowledge about which door doesn't contain the prize, the probability that we win from picking #1 is 1/3. However, we are given a knowledge: the host picked door #3 to show that it doesn't contain the prize.


We can use Bayes' theorem to determine the probability of winning if we switch to door #2:

P(door #2 has prize | host picked door #3) = P(host picked door #3 | door #2 has prize) * P(door #2 has prize) / P(host picked door #3).

By the rules of the game, the host couldn't pick my door (door #1). So P(host picked door #3 | door #2 has prize) = 1, since the host wouldn't readily reveal the answer by picking door #2. And, of course, P(door #2 has prize) = 1/3.


The hard part is figuring out P(host picked door #3). There are two possible scenarios: (1) the host picked door #3 when he knew door #1 contains the prize, and (2) the host picked door #3 when he knew door #2 contains the prize.

Again, using conditional probabilities,

P(host picked door #3) = P(host picked door #3 | door #1 has prize) * P(door #1 has prize) + P(host picked door #3 | door #2 has prize) * P(door #2 has prize)
= (1/2) * (1/3) + (1) * (1/3).


Put things altogether to find that,

P(door #2 has prize | host picked door #3)
= (1) * (1/3) / [(1/2) * (1/3) + (1) * (1/3)]
= 1 / (3/2)
= 2/3.


So by switching to door #2, you double the chance of winning.


----


We can similarly show that P(door #2 has prize | host picked doors #3-6) = 5/6. But prize = getting shot, so you definitely don't want to switch turns.

 

[luoapp]

^^^^two questions are different

 

[kgtrep]

^^^^two questions are different

Maybe. Here was my thought:

If I had entered the roulette not knowing anything (in particular, that the first four will not die), then the probability that I will die going last is 1/6.

But here, the first four didn't die. It's like, the gun is the host and it deliberately shot four other people besides me, because it knew the first four were blanks.

(This sounds really morbid, I know. We need to question Two Words' professor's morality.)

It then asks me do you want go next (switch doors) or go last (stay).

 

[boviscopophobic]

You and 5 other people are going to play Russian Roullete. You choose to go last. Persons 1 through 4 all survive their turn. Person 5 asks you if you want to switch turns. Do you say yes or no? Does your answer effect your probability for survival?

Do they re-spin between the trigger pulls? If so, you should go last (assuming the game ends when somebody "loses".) If they don't re-spin but the cylinder holds 4 rounds or fewer, then the gun isn't loaded and you can do whatever you like. If they don't re-spin and the cylinder holds 5 rounds then you had definitely better not switch, and if it holds >= 6 rounds then your chance of survival is the same whether you switch or not.

So I guess what I'm saying is, it is never better to switch.

 

[KharmaPolice]

In the roulette game it doesn't matter weather you change order or not, because the first four players have no additional information abou the game. If you want you could write down a probability tree.

The difference to the monty hall game is as follows (lets assume the game has six gates):
After choosing your door, the host will open 4 doors that are guaranteed to be empty (and thus giving you additional infotmation). But in roulette the first four shots are not guaranteed to be blanks. One of the first four players might very well have died. Thus no additional infotmation about where the bullet is among the remaining two shots can be derived.

 

[Gotchaye]

In the roulette game it doesn't matter weather you change order or not, because the first four players have no additional information abou the game. If you want you could write down a probability tree.

The difference to the monty hall game is as follows (lets assume the game has six gates):
After choosing your door, the host will open 4 doors that are guaranteed to be empty (and thus giving you additional infotmation). But in roulette the first four shots are not guaranteed to be blanks. One of the first four players might very well have died. Thus no additional infotmation about where the bullet is among the remaining two shots can be derived.

To put this another way, Russian Roulette can be understood as players taking turns randomly selecting one chamber from the set of all chambers. In Let's Make a Deal, Monty Hall randomly selects one door from the set of doors with goats. The extra information is being communicated to you by Monty Hall, and it's possible for him to be giving you information because he actually knows where the prize is.

Another problem which is fundamentally similar to the Russian Roulette one would be shuffling a deck of cards and then choosing cards to turn face up until the Ace of Spades is revealed. Given that you haven't found the Ace yet, the probability of finding it when turning any given card face up is just 1/cards_remaining. You don't improve your odds on the second draw by setting the first card aside face down, looking at the second card, and then replacing the third card with the first card and turning the third card face up.

 

[cpp_is_king]

You and 5 other people are going to play Russian Roullete. You choose to go last. Persons 1 through 4 all survive their turn. Person 5 asks you if you want to switch turns. Do you say yes or no? Does your answer effect your probability for survival?

If they respin between each turn then you should wait. If you wait you have two chances to survive: one when #5 spins, and one when you spin (because if #5 dies games over)

If they don't respin then it doesn't matter if you switch. It's in one of the last two chambers. If you go first 50% chance of dying, if you switch 50% chance of the other person surviving (which also means you will die)

 

[Two Words]

Yes, this was meant to be without respinning. I couldn't tell if this problem falls into the Baysian probability of prior and posterior probability. It seems like new information doesn't change the probability here. I suppose if it doesn't matter then I'd switch. I don't want to pull the trigger knowin a bullet is in there.

 

[boviscopophobic]

You do gain information. For concreteness, let's say that the cylinder has 6 chambers. You can label the chambers of the cylinder 1-6, with 1 being the chamber for the first player, 2 being the chamber for the second player, and so on. When you start out, you have some prior distribution on which chamber you think the bullet is in. If you have no strong reason to believe otherwise, a uniform distribution on 1-6 is a reasonable prior.

Next, each of players 1-4 probes their respective chamber. If the bullet is in a probed chamber, it is "found" with probability 1, else that chamber is empty with probability 1. After each player takes their turn, your posterior distribution on the bullet location becomes a uniform distribution on the remaining unprobed chambers. Thus your posterior probabilities do change. Furthermore, your subjective probability of death goes up as the game goes on, because the posterior probability of chamber 6 goes up after each of the players 1-4 takes their turn. Unfortunately, since your posterior distribution remains uniform (on the unprobed chambers) at each stage, you don't get any insight into which of the remaining chambers holds the bullet.

 

[Two Words]

You do gain information. For concreteness, let's say that the cylinder has 6 chambers. You can label the chambers of the cylinder 1-6, with 1 being the chamber for the first player, 2 being the chamber for the second player, and so on. When you start out, you have some prior distribution on which chamber you think the bullet is in. If you have no strong reason to believe otherwise, a uniform distribution on 1-6 is a reasonable prior.

Next, each of players 1-4 probes their respective chamber. If the bullet is in a probed chamber, it is "found" with probability 1, else that chamber is empty with probability 1. After each player takes their turn, your posterior distribution on the bullet location becomes a uniform distribution on the remaining unprobed chambers. Thus your posterior probabilities do change. Furthermore, your subjective probability of death goes up as the game goes on, because the posterior probability of chamber 6 goes up after each of the players 1-4 takes their turn. Unfortunately, since your posterior distribution remains uniform (on the unprobed chambers) at each stage, you don't get any insight into which of the remaining chambers holds the bullet.

I'm not sure if that was a mistake I made in my explanation, but that is what I meant.

 

[Pryce]

Hey guys. I could use some help.

So after graduating high school and after two or so years working full-time, I finally decided on a major (CS) that I hope to start next spring (I'm 21).

I was always pretty good at math. Had A's and B's in Algebra and all that, but it has been so long that I'm very rusty. I've spent that last 6 or so months brushing up on my algebra and it's going well but I completely fall apart when it comes to anything outside of the equations. As in, word problems (say, for example, math problems outside of what is on Kahn Academy).

Are there any books or websites or tips that anyone could give me? Am I just understanding the problems but not fully knowing how to apply them? I'm struggling here.

 

[cpp_is_king]

Hey guys. I could use some help.

So after graduating high school and after two or so years working full-time, I finally decided on a major (CS) that I hope to start next spring (I'm 21).

I was always pretty good at math. Had A's and B's in Algebra and all that, but it has been so long that I'm very rusty. I've spent that last 6 or so months brushing up on my algebra and it's going well but I completely fall apart when it comes to anything outside of the equations. As in, word problems (say, for example, math problems outside of what is on Kahn Academy).

Are there any books or websites or tips that anyone could give me? Am I just understanding the problems but not fully knowing how to apply them? I'm struggling here.

Not a lot of info to go on. Perhaps you could post an example of a question you struggle with.

When looking at the solution do you follow it?

 

[Two Words]

I have an interview for a tutoring job in the CS department of my university. The area where I am mostly shakey on is structural induction. I feel like I understand the concept of structural induction, but I sometimes have trouble applying it. Here is the way I see it. If anybody things my way of thinking is lacking let me know. I am also trying to find examples with strings as well.

Structural induction is done by showing that some property holds for all of the base cases of some recursive definition and when it is assumed to be true for some arbitrary amount of recursive steps, it can be proven that the next recursive step only produces things that have that same property. It is similar to mathematic induction. You could think of k being some positive integer that represents the number of time the recursive step is applied. My problem is applying this idea to show that a set defined in set builder notation is equal to a recursive definition. I know that I must prove that each are a subset of each other and am able to do it correctly, but I sometimes find myself confused on which way I am going. I can have trouble sorting the details that are important to what I am proving. Any good examples to help with this?

 

[Dereck]

I don't understand this finding the LCD thing for fractions.

If I have for example

-3 /x + 5

That means that I need to find the LCD, which is x / x

So that means I multiply both of these by x / x or some bullshit.

-3 / x &#8226; x / x + 5 / 1 &#8226; x / x

Which would leave me with this

-3 + 5x / x

or

- 3x + 5x / x

I don't understand, both of those are probably wrong

 

[Linkstrikesback]

First To get a common denominator of two fractions, multiply all the terms by every denominator of all the other terms.

In your example 3/x+5, the first part, 3/x has denominator x so we multiply only the other terms by x/x, so you now have

3/x+5x/x

As the second d term doesn't have a denominator (well, technically, it did, but it was just 1), we now have a common denominator.

The fraction is now (3+5x)/x.

I think that's already the least common denominator, as no further simplification can be made.

 

[Dereck]

If I have something like

4 / (3x+1) / (x - 2)

Is it OK to make it into this?

4 * (x - 2) / (3x + 1)

Which is

4x - 8 / 3x + 1