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Graphically, it is approaching -2.
Ok, how about this:
Substitution Time!
And put it all together and what do you get:
Well, look closely at the original equation. There must be a reason why it was written that way instead of as a straightforward quartic equation. It must be so as to make it clear that a substitution is meant to be involved so one should be looking around for a substitution instead of trying to solve the quartic directly.
I understand that, and agree, but these sorts of problems really upset me.
I know they shouldn't, and I know it's irrational, but I kind of feel like problems like these are gimmicky. I wonder how many times in a real-life situation will you come across a problem like that?
Your answer has a certain beauty to it, no doubt. It almost dances as it comes to its conclusion, (as G.H. Hardy said, "The mathematician's patterns, like the painter's or the poet's, must be beautiful"") But I feel like approaching like I did, has more value. I think in real life when you come across an equation very rarely will it be able to be solved with a nifty substitution.
Then again (and I'll contradict myself before someone else does), if I felt that way, why not just graph it and get a "good enough" approximation? Approximations are all you really need in real life.
I guess that's the difference between "Pure Mathematics" and "Applied Mathematics."
Or maybe I'm just jealous that your answer was cooler than mine.
Must I do everything around here all the time?
Oh come on, We pretty much had that one under control before you got involved.
Awesome answer though.
x = −5 is another solution.
Yeah, I saw that graphically, but I was hoping someone would come along and answer it analytically.
By the way, the other two roots are complex.
Using Pi man's x=1 and Synthetic Division I factored it down to:
(x-1)(3x³ + 27x² + 73x + 65) = 0
I'm curious: Pi Man, what method did you use to find that value?
Chain rule madness!
Not really worth LaTexing:
f((g(xf(x)))²) = f(h(x))
h(x) = g(xf(X)))² = k(x)²
k(x) = g(xf(X)) = g(m(x))
m(x) = xf(x)
Now, Working backwards I'm going to take the derivative of each function I defined:
m'(x) = xf'(x) + f(x) product rule
k'(x) = g'(m(x))m'(x) = g'(xf(x))(xf'(x) + f(x)) chain rule
h'(x) = 2k(x)k'(x) = 2g(xf(x))(g'(xf(x)))(g'(xf(x))(xf'(x) + f(x))) product rule
SO:
d/dx f(h(x)) = f'(h(x))h'(x) = f'((xf(X)))²)(2g(xf(x))(g'(xf(x)))(g'(xf(x))(xf'(x) + f(x))))
Well, that was a pain. With problems like these it's easier (for me at least) to color code them. Maybe I'll do that tonight, I have to go now.
PS I hope everyone looks at my work, THERE'S A LOT OF ROOM FOR ME TO MAKE A STUPID MISTAKE.
Numerical methods is basically just number crunching, the kind of tedious, repititious process that we normally leave to computers.
Well, that doesn't seem so bad. Maybe a bit useless, but not so difficult to learn.
Simply awesome. I wish I had the money to go to the exhibit.
I think that's something I don't understand. The other stuff is all stuff I've seen before, but in a bunch of different classes, Not just one.
With the exception of "numerical methods" which is simply a term I haven't heard before, those are all things I've covered before. Interestingly, in a few different classes, not one.
Can anyone give an example of a "numerical methods" question?
Yes, read this wiki on Riemann sums it'll give you an idea of how to go about doing it by hand.
Or you can plug it into a TI-83 or another Graphing program and evaluate it numerically that way.
Is this an integral for school, or for fun? Did you just make it up? Would an estimation work (like Riemann sums)?
Im an American, but might have taken the US equivalent to "Further Maths" What does it entail?
I don't think it can be done. I plugged it into The Integrator and it didn't give an answer.
Maybe the reason I haven't seen this before is because I've had minimal experience with vectors. In our class we would simply define a coordinate system and work with x and y. I know what a vector is use, but really didn't use them.
Thanks for clearing that up.
I'm a little confused... are i and j axis like x and y in the Cartesian plane? If so, why use i and j and not x and y.
When I took physics we used x and y. Is this because it was a high school physics class and you're in college? Or is it because I'm American? Or just because?
Not to sound like a teacher (I'm most definitely not one) But my rule of thumb is not to let a calc to the work unless I already know how.
In other words, I make a point not to be stupid, but I have no issues with being lazy.
Point made. I don't understand a word of that.
Positive, even/odd, divisible, functions, and relations are all very fimiliar terms, but I guess their meaning runs a little deeper than I realize.
Really? Your method works just fine for me. Where are you getting stuck?
Maybe I'm not fully understanding the depth of what you're saying, but I feel like all of this is stuff I already know.
Do you know of any (non-textbooks) books that would be a good place to start with intro to proofs? What's in an intro to proofs class? I've read many proofs (mostly famous ones) and they didn't seem to difficult to follow. Is there something more to it?
Any time I've had to tutor someone "from the beggining" I usually ask them where to start.
I ask them what grade they got confused in, then I just kinda do examples until I get a feel for what they know. I don't think there is any exact science to it.