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Each of us is of the form
, whereWe need to prove that b = 0
I definitely cross the x-axis at least once.....
How about you stay up all night and study math. Being amazing at math is way more important.
Yea ibgot 4.4ish using the graphing calc as well, but i was kind of dead set on solving it:( Im going to have to get some of the books bobbym is talking about. Evenetually. I am one who likes to know how things work. Im not satisfied unless i do.
I don't know why the book has me doing this.... I wasn't taught how to solve something like this yet. This is actually part of a calc 1 problem. Im graphing functions. It's part of the numerator for the second derivative. There's actually a 2 that I factored out, but it shouldn't be need to solve this. Well, technically now that I read the book it wants me to use a graphing calculator to find the important points. This polynomial is whats need to calculate possible inflection points.
I tried rational roots and it provided me with nothing that works, lol.
The derivative of sin x is cos x. Derivatives are equations used to calculate the SLOPE of a tangent line at some point on a curve. The curve is a function such as X^2 or in your example sin x. The derivative of X^2 is 2x. 2x is a linear equation and therefore is refereed to as a SLOPE equation. You can use that slope equation to calculate the ACTUAL slope of a tangent line on the curve of X^2 by choosing a point within the domain of the curve. Lets choose x = 2. The slope would be 2(2) = 4. The slope is represented with the letter m, so m = 4. Now that you have the slope, you need a y value to go with your x value of 2. Use that x value with the original equation to get yourself a y value: (2)^2 = 4. So now you have yourself a the slope of a tangent line at the point (2,4) on the curve X^2. Now what you need is the equation of this tangent line at that point. You remember the point-slope formula? Plug in all the information you have and solve for the equation for that tangent line: y - y2 = m(x-x2); y - 4 = 4(x - 2); y = 4(x - 2) + 4 = 4x - 8 + 4 = 4x - 4. So you have a point (2, 4); a slope equation of f'(x) = 2x; a slope of m = 4; and an equation of a tangent line, y = 4x - 4, at that point.
That's a basic derivative explanation. Forgive me if it appears that I'm assuming you know nothing.
It works the same for the function sin x. The slope equation sin x is f'(x) = cos x. One thing to keep in mind is that the graph of sin x goes to infinity to both direction. It repeats, goes round and round the unit circle. This is why your given an interval of (-pi, pi).
Now, it says determine when f is increasing and decreasing. A straight line is increasing when it's slope is positive and is decreasing when it's slope of negative. So if m = -2, that line is decreasing and if m = 4, that line is increasing. Positive increasing, negative decreasing.
So you have a function sin x and a slope equation for that function, which is cos x. Do you know you know how to find minima and maxima? The slope of minima and maxima is zero, m = 0. So now you know how to find minima and maxima. You set the first derivative, aka the slope equation for the function sin x, to zero. So now that cos x = 0, you have to ask yourself when is cos x zero? Cos x is zero at pi/2 and 3pi/2.
You might be asking why you have to find minima and maxima? By finding minima and maxima, you'll know where to check for increasing and decreasing slopes.
So now that you know that cos x = 0 at pi/2 and 3pi/2, are these values within your interval? pi/2 is, but 3pi/2 is not. If pi/2 is within the interval (-pi, pi), then so is -pi/2.
So you have these two points now. At these points cos x = 0 and being that cos x is the slope equation for sin x... the points on sin x with a slope of zero are the minima and maxima, the high and low points.
Now that you know where the high and low points are on sin x, those being -pi/2 and pi/2, you now know where to look for increasing and decreasing. You look between the high and low points, the minima and maxima. So you need a point between 0 and pi/2, and pi/2 and pi. Let's chose pi/4 and 3pi/4. If pi/4 and 3pi/4 are in the interval (-pi, pi) then so is -pi/4 and -3pi/4.
Now that we have points between -pi, -pi/2, pi/2, and pi; we can check for increasing and decreasing. Just use your calculator for this. f'(pi/4) = cos (pi/4) = 0.71, is that increasing or decreasing? Its increasing. Now we check f'(3pi/4) = cos (3pi/4) = -0.71, is that increasing or decreasing? It's decreasing. Do that rest and you we see whats happening.
Now remember! SIN X is the FUNCTION or CURVE. COS X is the DERIVATIVE or SLOPE EQUATION for a tangent line touching the function sin x. 0.71 and -0.71 are SLOPES of the tangent lines, the slope being the variable m, at the points pi/4 and 3pi/4.
If you want, you can solve for the equations of the tangent lines that same way I did in the very first explanation. Solve sin x using your x values to get y values. Using those y values, the x values, and the slopes, you can use the point-slope formula to solve for the tangent line equations.
I hope this helps. I can't believe I spent this much time writing all this out....
Thank you Bob. Yes, I mean maxima/ minima. The book I'm using refers to these points as critical points, sometimes extrema as well. I read through the section again and the book does say that there is extrema when f'(c) = 0 or f'(c) = dne. I guess being so tired I wasn't thinking straight. Also, after moving onto the next section it clearly states:
extrema when f'(c) = 0 or f'(c) = dne
and
possible inflection when f''(c) = 0 or f''(c) = dne
Thanks for your time bob.
I know that with polynomials I solve for the "zeros" of the first and second derivatives and I know I do the same with rational functions but... can someone please clear up where the zeros for each are solved? I was always a bit lost with certain things about rational functions.
Are critical points(extrema) solved in the numerator or both the numerator and denominator? I know that critical points are points where f'(c) = 0 or fails to exist so I'm thinking both the numerator and denominator?
What about possible points of inflection? Numerator or both numerator and denominator? My book doesn't say.
Ok thanks. Ill try this when i get home. I understand a lot of this but cant quite seem to fully wrap my head around it. Having trouble visualizing whats going on. For the most part i understand. Just need to spend more time working with it. Also, is cosine inversely proportional to sine? I noticed the behavior of the two when the triangle moves and it appears they are.
Thanks bob:) I do have a quick question though. Do you know of a quick way to figure out what quadrant a radian is in without using a calculator?
For example: -17pi / 6
Degree's is easy but radians is a bit challenging without a calculator.
I know I have to find a coterminal angle. I'll add 2pi which will give me -5pi/6, but even with the number looking a bit nicer, I'm still having trouble picturing where this lies within a circle.
I know I most likely need to find a reference angle, but I don't know whether to use pi or 2pi to find it.
I know the answer to this problem, but I had to use my calculator to find out. What is the trick to doing it without a calculator?
Thanks again.
Oh my goodness I think I understand this now. So stupid of me. I thought the radius was 3 initially. It's not. 3 is x and 4 is y. I solved the hypotenuse which is the radius. So sin theta = is 3/5. How could I not understand this before! LOL. Like losing your wallet and finding it after realizing it was on the kitchen table right in front of you the whole time. Thanks Bob.
The problem I have to solve is basically an angle in standard position on a Cartesian plane with a terminal end of (4,3). The problem wants me to solve for the six trigonometric functions. I know that I have to solve for the hypotenuse, which is 5. So that means sin=3/5 and so on for the remaining 5 functions. My question is, what about sin=y/r, cos=x/r, and so on? Im confused. So initially I thought I thought sin=3/4, because 3 is or could be the radius on a unit circle and y is... y. Is sin=y/r ONLY for when dealing with a unit circle??? Technically the problem doesn't mention anything about a unit circle. I don't know what is wanted and when or if they are the same thing?!?! I mean clearly 3/5 and 3/4 are not the same thing. I know that sin=opp/hyp and sin=y/r, but it seems to me that those are two different things. What am I missing here? thanks in advance.
You multiply both sides by I, oh OK. Before reading this I came to this:
... and you get where I'm going with that. Your way look more correct than mine though, because your way makes more sense than mine. Thanks.
I'm a bit confused here. I'm pretty decent with my Algebra, but there's something that is confusing me here. I have these problems to do in my electronics class that is nothing more than something for values that involves Ohm and Joules law. I'm given just enough info to solve for some value, but I must comprise a formula to do it, using what I know about these two laws. Im a bit confused with the algebra involved with assembling these formulas.
An example problem...
I have to solve for I, knowing that that:
The Laws:
V=IR
P=VI
This is what I did:
Now I come to this point...
... and naturally want to do this:
But this solves for voltage not current. After messing around with it some I learned that it's:
This confuses me:( What's the deal here? There was another scenario where I did the same thing and the result was wrong.
Is it because this:
and this are not the same?
Because I would technically be replacing the V with a 1? Is that wrong? Well, I mean it's wrong because it obviously doesn't work. I remember from Algebra 2 with complex fractions that a 1 was added in there somewhere and then the fraction were reciprocated and then multiplied.
Hi;
It is incorrect to always seek an algebraic method as a solution for every type of equation. When there is one we are happy but they are rare. The hard fact is most equations can not be solved by algebra, usually iteration is used and this starts off with a guess!
Ugh, that drags, but there is always a way to solve something. An official method. Instructions. I know that guessing is a thing with mathematics, but there's always an explanatory way of doing the math. Thanks for you help. I fully understand this now. Got more than half of my homework done early early this morning, and it only took me about 45 minutes once i had it down good.
I think I'm starting to understand this. I had to skipp ahead several sections before it actually started to teach how to do this. What a very poor way of teaching. Frustate the hell out of the student as a sick joke before teaching. This is more what I was looking for:
I guess I was looking more for an algebraic way. Not that what you were showing me wasn't algebraic, wanted to see it worked out like this so I can get a clear mental picture of whats going on. I was literally doing the problem for section 4.4 before the problems in section 4.2. Can you believe that? Why would they do that? Just horrible.
Thank you for your help, but It still looks like something I would have to trial and error to solve. Meaning no easy way to do it. I skipped ahead in my book to see if there was anything it was going to teach me that I could use in the earlier section. I saw the changing of the log formula, which is what you are doing, but still... what power do I have to raise 8 to get 16 or whats the root of 16 to get 8? That's it. You kind of just have to play around to get the answer. Its very fortunate the 8 and 16 have the same base of 2, 2 being an even more helpful base to use, but what if two numbers don't so easily have the same base? I mean you can give them the same base but... I don't think it helps for many. This really drags. I did so well up until this point.
The book I'm using for class drags! It was OK all the way up until chapter 4. Section 4.1 was about natural e and it was a bit blah explaining natural e but I think I know about e for now. Section 4.2 is where I am at now and I just don't get it. It's about Logarithms. I understand the concept but there is no mathematical or algebraic explanation on how to solve it. In the book and even this website basically tells you to just take a shot in the dark or scribble the hell out of a sheet of paper to find the exponent. For instance, on this site, the example is:
The same with this example:
The explaination:
WELL, 1/8 = 0.125
Well what?!?!?! What if I didn't know that? They are not all that easy! Here's one from the book that I have been trying to figure out a method for:
I know the answer is 3/4, but do I solve to get that?!?!? Other than using trial and error. This is as far as I got to understanding how to solve for it:
I see a 4 and a 3. This makes me think I'm close, but I I can figure out how it's solved for 3/4. Can someone please explain to me how to solve these problem without using shot in the dark trial and error? Thank you. I have been working on this for 2 days now and I'm tired. I have to move on but I don't understand this.
That was easy! I found another math book online that explains it better. BUT! I still have no idea how my math book is trying to do it! This online book had me replace N with MR, where my book is does M=N/R. I understand why the book does that, but I can't figure out how the R ends up in the exponent and the book doesn't explain why.
I understand why M=N/R, because:
I don't understand why R was added to the exponents. I mean I do, because we need to specify a rate. I guess what I mean is, how was it added mathematically? I have been trying to figure this out all day. For the most part I understand it, except for that.
I'm learning exponential functions. I'm trying to understand this. For the most part I do, but somethings just... it's so weird, but I I'm confused as hell. For instance:
I just want to simplify this. LOL. I don't mean this either:
I think it's this:
It looks retarded though. Is this right? I checked it using my calculator and I get the right answer but it looks weird. They have different bases and different exponents so there nothing more you can do with them.
Let's say the 3 was a 2:
So:
Is that right?
Kronecker wrote:God created the integers, all the rest was the work of men.
LOL.
I just have to say again that I feel so darn stupid. I don't know why I didn't see this before. I think I was stuck on everything being an integer.
I do know how to do long division but the leading coefficient is larger than the leading coefficient of the numerator. You can't divide
by . What do I need to multiply with to get ? An x, but what about the 2? You can't divide 1 by 2. OH HELL. LOL. HA HA HA HA, that's funny. I just figured it out. LOL. Typing it all out in english is what cleared it up. WOW. I feel so stupid. HA HA HA.I'm so frustrated. The book I'm using for school doesn't show me how to do these, yet asks me to do them! I checked online for help and the same type of problems the book explains, is what everyone online to explaining. Please help.
The problem:
The answer:
I have no idea. The leading coefficient of the denominator is greater than that of the numerator. Can't find anywhere on how to do this! I am wasting so much time trying to figure this out. Please help.
Hi;
The rational root theorem only provides a possible list of rational roots of a polynomial. They do not have to be roots of the equation. 5 / 2 is not a root of that equation so it will not work as you found it.
There are other methods to locate and isolate the roots.
OK. Thanks. I haven't been taught how to solve higher degree polynomials yet, other than the rational roots theorem, so I will not go here just yet. Don't want to veer of course. The problem I was solving didn't ask me to factor this polynomial. This polynomial is the numerator of a rational equation. I was asked to determine the domain, horizontal asymptote, and the vertical asymptote. I kind of veered of course with trying to factor it.
Thanks.