1/3x^ = 2x+1

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

What's your x (on the left-hand side) to the power of?
2
You need to remove any x's you have from the denominator. So, multiply everything through by 3x^2 (I said 3x^2 instead of just x^2 because we'd end up multiplying through by 3 anyway - may as well get it over and done with). So,

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

\[\frac{1}{3x^2}=2x+1 \rightarrow 1=3x^2(2x+1)=6x^3+3x^2\]
he may have meant (1/3)x^2...
Is that what you meant, mac?
that would sound more likely to be the homework assigned to an algebra class
Because solving 6x^3+3x^2=1 algebraically is not nice.
^not at all...
\[\frac{1}{3}x^2=2x+1 \rightarrow x^2=6x+3\] after multiplying through by 3. You have a quadratic to solve:\[x^2-6x-3=0\]
ergh... that still wont give integer roots
You can complete the square or use the quadratic formula.
Can't help that...
\[x^2-6x-3=0 \rightarrow x=3 \pm 2\sqrt{3}\]
as for the 6xcubed i would use long division... or rather synthetic... is that the best way?
Do you mean you would use long division or synthetic after finding a factor?
i would use them to check if i can guess a factor lol
Yes
i wonder where mac went...
The method for solving cubics is a pain, so I usually try to find one solution and, like you said, use long/synthetic division on the cubic and go from there. Else, numerical solution, like Newton-Raphson method.
Mac's lurking...
newton-raphson method? whats that?
It's a method for finding solutions to equations when you can't do so algebraically (or can't be bothered). The only condition is that the function you're looking at has to be continuous (because you use calculus).
im curious, as the derivative cant tell you solutions... can it??
You're right, the derivative itself won't give you the solutions (unless you have a multiple root), but the method takes your guess, then uses calculus to find the equation of the tangent at the function value of your guess for x, then seeks out where the tangent cuts the x-axis. That point - where it cuts the x-axis - is closer to the root than your initial guess (usually...there are cases where it won't work). That new x-value is your new 'guess' which goes through the process again to yield a better 'guess' for x...you can keep going until you get to a level of accuracy you want.
... so this will only work with f(x) with a degree 3 or more so that f'(x) is at least quadratic... and you find the tangent, then find the zero of that equation, and then use that in the original and find the tangent and use the zero of that equation and so on?
The method will will work with any function that has at least one derivative. I'll try and find some information for you. It's a very useful tool to have, mainly because you end up having at your disposal a means to solve many practical problems that either don't have algebraic solutions, or where the solutions are a pain to get at and you don't need your answer to be exact.
but then you will never get the exact zero, you will only get closer and closer to it? as the zero of a tangent line can never be equal to the value put into the original function for that tangent line unless you have a multiple root case ( as you said above)
Check out the image - this is what's going on. http://en.wikipedia.org/wiki/Newton's_method
thank you man
You're welcome ;)
what the heckkk that is freaking genius. damn newton -.-
Yeah, I think it's his best thing outside of the calculus and Laws of Motion.
when it comes to physics laws of that sort, id always put kepler over newton... idk why newton got all the fame for those, keplers were way better -.-
Try it with something like finding the square root of 2 - it will find it pretty damn quick. \[x=\sqrt{2} \rightarrow f(x)=x^2-2=0 \rightarrow f'(x)=2x\]Then\[x_{n+1}=x_n-\frac{f(x)}{f'(x)} \rightarrow x_{n+1}=x_n-\frac{x^2_n-2}{2x_n}\]
i didnt understand the second half of that lol
I'm going to do an example.
You have to find something to start with, so we make an educated guess for sqrt(2). Since \[1<2<4 \rightarrow \sqrt{1}<\sqrt{2}<\sqrt{4} \iff 1 <\sqrt{2}<2\] as sqrt function is monotonic increasing (i.e. the direction of the inequality is preserved). So we have for an initial guess for x,\[x \approx \frac{3}{2}\]Then, a better 'guess' for x will be,\[x_2 = \frac{3}{2}-\frac{\left( \frac{3}{2} \right)^2-2}{2\frac{3}{2}} = \frac{17}{12} \approx 1.416667\]and the square root of 2 is about \[1.414213562\]
You can see after one trial, it's pretty close.
The closer you are to your answer in the first place, the faster it will converge. You could then plug 17/12 into the formula to get a better approximation, and continue.
Next one would be 577/408 ~= 1.414215686
dang, the formula, has you divide the tangent line function by the tangent line slope, why?
Well, going back to the sqrt of 2, when you start out, you don't know what the sqrt of 2 is, but you know it should be something, so call that something, x. Then x =sqrt(2). If I square both sides and subtract 2 from both sides, I get x^2-2=0 So, finding x = sqrt(2) amounts to finding where on the x-axis x^2-2 will cut it. So I plot my parabola, and imagine the setup that's in that Wikipedia pic. Since the method finds where the graph cuts the x-axis, it will find that x where x^2-2=0; that is, the x that came from x=sqrt(2).
yeah i get that, but i dont get how subtracting the tangent line formula divided by the slope of the tangent gets you the intersection of the tangent line?
The reason for why we divide the function by the derivative is due to the derivation. The derivation can be done in either a geometrical setting (using that picture, say) or with Taylor series expansion on an arbitrary function. I'm sure proofs would be online. It's a pretty old theorem ;)
oh wait i get it...
You know, you can derive it yourself by analyzing each step in that pic. and doing what the mathematics would dictate in that situation.
your taking y=mx+b and doing 0=mx+b so mx=-b so x=-b/m and you take that, which is the distance from the x value to the zero, and subtract it from your original point which will show you where on the x axis it is...
nope that made no sense... ill figure this out, gimmy a sec
You're half there.
It's difficult to explain online. They need to put a drawing feature in this thing.
lol they do, like tutorvista
I think mac must think his thread's been hijacked.
i think he left
so youre taking x and subtracting y/y'(x)
i still dont get why ><
Hang on, I'll see if I can put a small proof together.
how is the y value divided by the slope at that point, equal to the difference btween x and the intersection of the tangent line of x
no you dont need to make a proof of it i just need to understand how y/y' is the difference between x and the zero of y'
Wait, if I prove it using the geometrical route, you'll see...I'm just going to see if it's feasible (versus finding something online).
ok :/
Alright, writing it up isn't hard, but you'll see what's going on if you do a drawing like the one seen in Wikipedia.
And go along with it...
i can understand how it all works, but how does the drawing explain how the formula works?
(i already made a drawring btw lol)
Suppose x_1 is close to a root of the equation f(x)=0. Then the equation of the tangent at x=x_1 is\[y-y_1=m(x-x_1)\]that is,\[y-f(x_1)=f'(x_1)(x-x_1)\]This tangent crosses the x-axis at a point where x = x_2 and y = 0. Hence,\[0-f(x_1)=f'(x_1)(x_2-x_1) \rightarrow x_2=x_1-\frac{f(x_1)}{f'(x_1)}\]
Then repeat...
oh wow that makes sense...
Happy now?
yes :D thanks for spending time to teach me this stuff :O
No worries - it's good you want to know :)
its incredible that you can put up with teaching me like that, even alot of teachers would loose it pretty soon
i honestly am really curious as to where mac went...
he stopped observing a while ago even
You learnt quickly - it wasn't a problem. You would have learnt quicker if the interface was easier (but it's not the designers' fault - where's the holography already?).
Yeah, he probably got his answer and was all, "WTF is going on?!"
whats after differential equations?
Eh?
like course wise, you take calc1 calc2 calc3 diff eq then what?
Have no idea man, not American ;D
but you studied in europe, so the course order must be nearly identicle, especially in advanced mathematics, so instead i should probably ask, what did you do (courses) for you degree and btw, you have a masters? phD?
your degree*
Partial differential equations is usually end of the line for calculus. Then there's other stuff like Lebesgue integration (overtakes Riemann integral), and other stuff in algebra and geometry - the further you go, the more the areas start to link up. E.g. differential geometry, algebraic topology, etc. But for calculus, numerical and algebraic methods for solving partial differential equations is usually end of line. If you know how to solve (methods to solve) nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients, you're done!
Since i did not understand any of nth order, non-linear, inhomogeneous partial differential equations with non-constant coefficients i guess im not done :D
lol, keep going, though!
do you know all that is taught of mathematics to the extent that research is your only option?
No, you can do whatever you want with it. I teach as well. You can use it in industry and government, as in modelling for things...haven't really sold it, have I?
well not in that sense, i meant like if you wanted to know more on math, your only option would be to study unsolved theorems or unproved research or do research on your own
anddd man i need help on something: \[\int\limits_{}^{} (\sin x - \cos x) \div (\sin x)\]
nevermind i got it
Yes, you're right on your last point.
thats pretty crazy, that you know everything O.o
I don't know everything..!
you know everything in math...
No I don't ;)

Not the answer you are looking for?

Search for more explanations.

Ask your own question