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Fizban140
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Just got my first test back...anyways I am going over it and I don't understand some of the stuff on here at all. It isn't in the book and I can't easily find it online.
What is the domain of f-g
f(x) = square root of x
g(x) = 1/x-1
This looks really simple to me, can't divide by zero and can't have negative square roots, that makes sense to me...so I did that.
The domain should be 2-infinity, but that is wrong and I don't know why.
Also I can't find in my book on how to use () and [] to define what is used and what isn't, in class we use that but in the book they use x is not equal to n.
Find the x intercepts f(x) = -3x^2 + 5x
Pretty simple to me, I have done this same problem so many times. 0=-3x^2 + 5x
5x= 0 -3x^2 = 0 0 for the intercepts correct?
Next one has a graph, what is the range I put (negative infinity to 3) the answer was suppose to be to 4] What the hell is the difference?
Next question is what x is g(x) > 0 now let me translate this to enligsh....what x coordinate in y > 0
Wow what an easy question....just let me write down where y is greater than 0 on x...turns out I was wrong and the answer was two numbers that are 0 for y, I don't get it.
Next I had to write down where the graph was increasing, what an easy test...so I write down the coordinates where it starts to increase just like I did for all the homework...and I am wrong. Why? Is there some different language of math I have to use or is the graph not increasing when it goes up?
Next I had to sketch the graph with the peicewise determined function thing, I am quite good at this I thought since I spent about 8 hours on the homework but I won't even post the problem here, too difficult. I just don't understand these at all I guess.
I had to sketch a graph step by step, I guess I forgot the order to do it in so I got it wrong or something.
This one I was very confident on, if (4, -6) is a on the graph of f what point would be on the graph of...
d) f(4x) wow what an easy problem, got this for sure. (16, -6) how could that be wrong?
I am considering dropping this class, I don't think there is enough time in a day for me to review basic math, this chapter and then study for the next chapter for the test on monday. I mean just thinking about how much time I put in to fail this test I would have to at least double my time spent studying to catch up and then double the time spent studying for the next test, there isn't enough time. Now factor in my other classes and I am so fucked.
What is the domain of f-g
f(x) = square root of x
g(x) = 1/x-1
This looks really simple to me, can't divide by zero and can't have negative square roots, that makes sense to me...so I did that.
The domain should be 2-infinity, but that is wrong and I don't know why.
Also I can't find in my book on how to use () and [] to define what is used and what isn't, in class we use that but in the book they use x is not equal to n.
Find the x intercepts f(x) = -3x^2 + 5x
Pretty simple to me, I have done this same problem so many times. 0=-3x^2 + 5x
5x= 0 -3x^2 = 0 0 for the intercepts correct?
Next one has a graph, what is the range I put (negative infinity to 3) the answer was suppose to be to 4] What the hell is the difference?
Next question is what x is g(x) > 0 now let me translate this to enligsh....what x coordinate in y > 0
Wow what an easy question....just let me write down where y is greater than 0 on x...turns out I was wrong and the answer was two numbers that are 0 for y, I don't get it.
Next I had to write down where the graph was increasing, what an easy test...so I write down the coordinates where it starts to increase just like I did for all the homework...and I am wrong. Why? Is there some different language of math I have to use or is the graph not increasing when it goes up?
Next I had to sketch the graph with the peicewise determined function thing, I am quite good at this I thought since I spent about 8 hours on the homework but I won't even post the problem here, too difficult. I just don't understand these at all I guess.
I had to sketch a graph step by step, I guess I forgot the order to do it in so I got it wrong or something.
This one I was very confident on, if (4, -6) is a on the graph of f what point would be on the graph of...
d) f(4x) wow what an easy problem, got this for sure. (16, -6) how could that be wrong?
I am considering dropping this class, I don't think there is enough time in a day for me to review basic math, this chapter and then study for the next chapter for the test on monday. I mean just thinking about how much time I put in to fail this test I would have to at least double my time spent studying to catch up and then double the time spent studying for the next test, there isn't enough time. Now factor in my other classes and I am so fucked.
Fizban140 on
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On the intercept problem, there is more than just 1 solution. An easy way to see this is to factor out an x from your quadratic. F(x) = x(-3x+5). Where does that = 0?
Second one you need to use the quadratic formula to find the zeros or factor it, where the factors are x and -3x+5, then find where each of those are equal to zero (0 and 5/3)
The difference between 3 and 4 is the difference between right and wrong, apples and oranges, g&t and SE++
The 4x one, you had to find the equation of the graph, then plug 16 in place of x and find what y was. If you increase x why will change unless it's a horizontal line
the difference between () and [] is easy to remember - just think that the [] grabs onto the number and () cant grab it.
so the interval (-2,1] does not include -2 but does include 1.
always use () when it comes to plus or minus infinity.
on the first question is that f minus g?
as for the x intercepts, having a hard time following you. What were the values of x you got in the end - did you get the question wrong?
as for when a function is increasing you need to express it in terms of intervals.
graphing a piecewise function usually is determined by the intervals. If you made a mistake, i would check the endpoints.
as for the question if (4,-6) is on the graph, what other point would be on the graph did you have f? were you told anything about f? is it linear? essentially this says f(4) = -6, and you cant know much of anything without f or at least some facts about f.
As for dropping the class it really comes down to how you feel about this after working on it for a bit. It took me a long time to really learn how to study for math exams. A good indication would be if you feel a lot of it is learning how to refine your answers, tighten up your notation etc, you can probably do well. If when you see the full solution for one of these problems and it contains steps that you do not understand / ideas you are not comfortable with, this is where you need to examine how much review you need. Sloppiness is quite fixable, and not to be rude but i can tell in just the way you presented these problems that you are still learning to articulate mathematically.
lack of understanding of mechanics becomes a tougher judgement call.
Finally, for the f(4x) question, you did it backwards. If the equation was y=x-10, you would get a point at (4,-6) if the equation was y=4(x-10), the x input of 4 would give you a y of -24.
http://en.wikipedia.org/wiki/Quadratic_equation
This is a bit overkill, but some things here should look familiar. namely this should ring a bell:
If you haven't learned the quadratic formula you just have to factor then
-3x^2 + 5x = 0
Factor out an 'x' from each term to get
x ( -3x + 5) = 0
so then you can break this into two equations
x = 0 and -3x + 5 = 0 and find both roots, by solving for 'x' in each equation
so like veevee said you should get the roots as x = 0 and x = 5/3
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For a quadratic equation y=ax^2+bx+c use the equation (-b plus or minus the square root of (b^2-4ac) ) all devided by 2a. The + path will provide one place where f(x)=0 and the minus path will provide the other value of x. Then if you need the y value you just plug those x values into the original function
I mean all this studying that I did for this test, I don't even remember how to do most of this stuff and the test was not even two weeks ago.
Do you have a tutor yet?
You need a tutor.
There is absolutely NOTHING demeaning about needing extra help to understand something and learn effectively. I have in my career as student and engineer been both tutor and tutee and I can tell you straight up that personal attention is the most valuable instruction you will receive.
Everybody can do math. Everybody. No exceptions. Some of us just need more help than others in particular areas. There is nothing wrong with asking for help when you need it.
The greater detriment here is being too proud to seek out and accept help when you need it. Any tutor worth his salt won't be aiming to make you feel dumb. Tutors are there because they want to help (or make money _and_ help).
You have to get over this ASAP. My undergrad was in mathematics and I abused the hell out of my instructors office hours, as well as getting help in the Math Lab when needed. It is not a matter of being "stupid" simply reenforcing concepts outside of a class setting where the ability to absorb the material at your own pace just isn't there.
I guarantee that you are not the only one who feels this way. I had always wondered why folks in my classes were just taking notes and NEVER asked any questions - its not because they just "got it" (well I'm sure some did) - they were just more focused on getting the notes down, and were saving understanding for later.
Is this a Business Math class or Pre-Calculus?
I'm just going to quote this because Usagi has said everything that needs to be said.
You said you got help every Thursday. That is great, and you should definitely keep going. The other thing you should be doing is going to see the professor during office hours. Make sure you do it right though.
If you have assigned homework:
Do the work to the best of your ability. Do it early so that you can take that to your professor's office and say "I don't think that this is right, but I can't find where I've gone off the rails. I know this first step is this and then I need to do this other thing... if I am messing up can you show me where?"
If you do not have assigned homework:
Most math texts have problem sets with at least half of the answers in the back. Make yourself a set, work the problems, and then compare your answers to those in the book. When/if they don't match take that to your professor and ask what went wrong.
Either way, it is paramount to show your work. This way the professor can show you if a step is being done incorrectly and hopefully set you on the right track. This will also give you opportunities to ask things like "what is that" when the prof tells you to use the quadratic equation. And just like with tutors, no decent teacher will laugh at you or brush you off. They'll show you the equation and explain how it's used. They maintain office hours to help you understand that material better and get clarification on things that didn't feel quite right in lecture or that you got fuzzy on after being away from the course for a day.
Try this to supplement your instruction.
Once enough pieces fall into place, you'll start understanding more and more by yourself.
There's probably only a few things that you're not getting, and once you do, the rest will start making more sense. The reason for this is that math is all about logic, so it all hangs together. Notation may be arbitrary, but the rest all just fits together logically.
One quick note to the others here (to the OP -- only read this once you understand the math you were asking us about, otherwise, skip the following, to avoid any confusion) : all the math I see in the OP I learned in my last year of high school. The only significant difference I see is in the intervals notation: where you write (-1,3] I learned to write ]-1,3]
We would only use square brackets for intervals, no parentheses (to avoid confusion with the parentheses in functions, for example). If the bracket points inwards, the number is included, and if it points outwards, the number is not included. Also, if one side of the interval ends with +/- infinity, there is no bracket (because a bracket denotes the "end" of the interval.)
Anyone else here learn interval notation like this?
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There's two ways to proceed:
A: Recognize it's a skill like any other and get it slowly and carefully because you're not born with a talent for it (talent is not capacity; you can do a thing without talent, you just have to work harder at it)
B: Stick with the tutor
The thing about a tutor is that they aren't a distant instructor or a clueless search engine. They watch as you struggle with a concept and can instantly shift to another way of teaching until something clicks. The feedback is instant, and they can almost make up for a lack of talent.
I once knew someone over the Internet when I was 18. We were talking about a programming language I was learning.
"I wish I could learn that, but I'm too old to learn much."
"How old are you?"
"18"
"We're the same age."
Genuinely, honestly, and from painful experience, the only thing that can hold you back from learning math is yourself. There's no grand genetic lottery on ability to learn certain skills. We all win that jackpot.
I flunked remedial math twice before acing it.
A tutor is hands-on and can actually understand your misunderstandings.
Get a tutor!
It's much easier to communicate math in person, on a page, than over the internet. This is particularly true at the level you're at (no Latex, no Maple). I can't read alot of your post mathematically because you're throwing in what amount to non-sequiturs.
Get a tutor, and persevere. This material is completely manageable.
Finally: "Do not worry about your problems with mathematics, I assure you mine are far greater." Albert Einstein
Doesn't matter where you are in your math education... the stuff is hard.
Before I begin I want to make a note on the distinction between real numbers and integers. I think you know that already, but it's important for later on so just to be safe. The real numbers are basically zero plus any positive or negative number which contains either or both of a whole and fractional part. 0, 0.4, 5.2286, -26.5345, and -3 are all real numbers. An important thing to remember is that the real numbers are continuous, so that any two real numbers which are not equal to each other have an infinite quantity of other real numbers between them. Integers meanwhile are any real number with no fractional part. 0, 1, 2, -1, and -235463 are all integers (as well as real numbers, since all integers are real numbers too). Unlike real numbers integers are not continuous; if the difference between two integers is 1 then there is no integer between them.
I'm addressing this first since it'll be helpful later on. Both (a,b) and [a,b] include all the real numbers that are both greater than a and less than b. Think of the difference as how they treat the corner cases a and b. [ and ] have corners and therefore include the corner case, while ( and ) do not and exclude the corner case. So [4,6] includes the values 4 and 6 while (4,6) does not, and both include values such as 4.0000001, 5.2, and 5.9999999. When the two types are mixed just look at each side independently. [4,6) includes 4 but not 6, while (4,6] includes 6 but not 4. When a or b are -infinity or infinity we use ( and ) by convention; since infinity is a concept rather than a number it wouldn't make sense for a range to include it, so the parenthesis suggest everything "up to" infinity.
You should generally assume that the domain of a function is some subset (ie part) of the real numbers. In particular, its domain will be (-infinity,infinity) minus whatever ranges or values for which the function is not defined.
The limed statement is perfectly correct for real numbers (I'm ignoring complex numbers and you should too). The negative square roots part means we can chop off all the negative numbers from our domain, leaving us with a domain of [0,infinity), which is all positive real numbers plus zero. The dividing by zero part means that you correctly excluded 1 from the domain. However, remember that the domain is the real numbers rather than integers. That means that 1.000001 is still valid, as are 1.564, 0, and 0.4543 (for example). So what we want is the domain [0,infinity) excluding the single point at x=1. We can do that by combining two domains as such: "[0,1) U (1,infinity)", where that U is the union sign and basically means any number in either of those domains. So, in english, that says "Any number that is (greater than or equal to zero AND less than one) OR (greater than one)."
The key is to find values of x for which the entire equation equals zero, rather than finding where individual additive terms are equal to zero. In this case solving for the additive terms happened to partially work because both terms were equal to zero at x=0, and since 0+0=0 the whole equation was equal to zero too. However, that won't work in general (either giving you the wrong answer, omitting one or more correct answers, or both), and even in this case it doesn't find the x-intercept at x=5/3.
So, how do we find the places where the entire equation equals zero? Other posters have mentioned the quadratic formula, and assuming the equation is actually quadratic (more on that in a bit) that formula will give you the answer(s) if they exist. Technically it will always give you an answer, but it may be a complex number, and like I said before we're ignoring those, so sometimes there won't be a (real number) answer.
But, the quadratic formula is a little ugly looking, so before we get into that are there other ways to find the answer? Yes, maybe. This is where factoring comes in. Factoring sometimes allows us to take an expression (where an expression is anything from a simple constant number like 2 to any complicated thing like 3x^3+2x^2-9x+7) and break it down into a product of two or more simpler expressions which is equivalent to the original expression. We can then evaluate each simple expression to see where it is equal to zero, and then that will be a zero of the entire equation.
"Wait a second!" you may be saying to yourself, "didn't you just say a few paragraphs ago that you can't do that?" The difference is because we're multiplying terms rather than adding them. As an example, let's say that f(x) = (expression 1)*(expression 2)*(expression 3). If expression 1 is equal to zero when x=2, then 0*(expression 2) will be equal to zero because zero times anything is zero. Likewise 0*(expression 3) will equal zero, and so f(2) = 0. However, with addition, if f(x) = (expression 1)+(expression 2)+(expression 3) then expression 1 being equal to zero at x=2 doesn't guarantee the whole equation will be equal to zero like it would with multiplication.
Now, even factoring can be a bit tricky, and this post is getting pretty long so I'm hesistant to go into too much detail. But, for the equation above notice that both additive terms have at least one x in them. Due to the distributive property that means we can divide each term by x and then multiply the whole equation by x to avoid changing the value, thus going from -3x^2+5x to x*(-3x+5). Now we have an equation with the form (expression 1)*(expression 2), which means we can set each expression equal to zero in turn and find the zeroes of the whole equation. The first expression is simply x, which is zero at x=0. The second one is -3x+5, which we plug into the equation -3x+5=0 (since we want to find out where the expression is 0). We add 3x to both sides (since the -3x term is negative) to get 5=3x, and divide by three to get 5/3=x. So, the zeroes of f(x) are x=0 and x=5/3.
Factoring can be quite useful, but it can involve trial and error and won't work for every quadratic equation. So, let's look at the quadratic formula. First, the quadratic formula is actually two separate equations that are normally written as one pseudoequation because their forms are very similar. I suggest going to Wikipedia to see it visually, since writing it out on the forums is a bit messy. I'm going to work through an example anyway, but seeing a clean equation will make it simpler to follow along. Second, in order for the quadratic formula to work you need an expression of the form ax^2+bx+c=0. a, b, and c will most often be non-zero integers (including negative numbers), but they can be any real number including zero and the equation will still work. Note that if one of them is zero the corresponding term will almost always be omitted; the equation -3x^2+5x=0 is equivalent to the form -3x^2+5x+0=0 and so you'd have a=-3, b=5, and c=0.
(Two notes: sqrt() is the square root function, and PLUS and MINUS are normal addition and subtraction; I just wanted to make the difference between the two equations obvious)
One half of the formula is (-b PLUS sqrt(b^2 - 4ac))/(2a). Plugging in our values we have (-5 PLUS sqrt(5^2 - 4*(-3)*0))/(2*-3) = (-5 PLUS sqrt(25))/(-6) = 0/-6 = 0.
The other half is (-b MINUS sqrt(b^2 - 4ac))/(2a) = (-5 MINUS sqrt(5^2 - 4*(-3)*0))/(2*-3) = (-5 MINUS sqrt(25))/(-6) = (-5 - 5)/(-6) = (-10)/(-6) = 5/3.
Hopefully my explanation above covered this, but in particular the second domain covers values such as 3, 3.3345, and 4 while the first does not.
If g(x) = 1/(x-1) from before, then g(x) is positive whenever (x-1) is positive, which is true whenever x is greater than 1. Could you clarify this part for us?
I suspect what they were looking for was the range of values of x for which the graph was increasing, rather than the coordinates where it begins to increase. So if I understand this right, if the equation was decreasing from (-infinity,0), increasing from (0,5) and decreasing from (5, infinity), you wrote 0 and they wanted (0,5).
This one's rather tricky. The easiest way for me to think about this is to think about x as a unit of time, perhaps x hours after some initial time (when x=0). f(4x) won't behave in a fundamentally different way than f(x), regardless of how f(x) is defined. The difference is how "fast" the function goes. f(4x) will have the same value at x=2 as f(x) has at x=8, and f(4x) will have the same value at x=3 as f(x) has at x=12. Basically the 4 in f(4x) makes it "go" four times "as quickly" as f(x), and so f(4a) will equal f(b) whenever b/a=4.
Going back to the problem, we want to figure out 'a' when b=4. b/a=4, plug in b and get 4/a=4, then we multiply both sides by a and divide both sides by 4 to get 4/4=a, and so a=1. That's the x coordinate we need, and since f(4a) = f(b) and f(b) = -6, that means our point is (1, -6).
Make sure you're attending lectures and taking good notes, do all the problems you can find, circle every single one you have trouble with, and then take them to your teacher's office hours. That's what s/he has them for. If you're having a lot of trouble and aren't doing this you're basically sabotaging yourself.
Man the fuck up. Having a shit attitude won't help so just stop it.
Get extra help. I'm working on a graduate degree in a highly mathematical field and still have to go to office hours to understand shit. I prefer to work things out on my own, but guess what... sometimes it just doesn't click. Swallow your pride and get as much help as you require.
1) You are fully capable of doing this. Math is a hijacking of mental constructs which were not meant to perform the tasks we're requesting of them so it will not always be easy, but you can do this.
2) You need a tutor. I don't mean the kid in the math lab working on their teaching degree, I mean a professional who will be able to see what you are missing just from watching you work a problem and then be able to explain it to you in a way that will make sense to you. This is not necessarily a role that your professor can fill. Many professors are terrible teachers because the ability to communicate ideas effectively is a separate skill from being able to do math.
3) I was that kid working on his teaching degree for a few years there and I can tell you that most of the people taking remedial math had no other problem with math than that they thought they couldn't do it. Get that thought out of your mind fast. You are fully capable of creating a mental block here that will make this a hundred times harder than it needs to be.
4) I am now the professional tutor that I was talking about earlier and I can see several problems in the way you're looking at this. It would take pages and pages to explain it all in text or a few hours over a couple of weeks in person. I don't like to broadcast my location over the internet but if you're somewhere in the RTP area I may be able to help you. Send a PM if you're interested.
0431-6094-6446-7088