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#1 (permalink) |
The GrandDaddy of them all!
Location: Austin, TX
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Limits
I'm lost at how to do this problem.
![]() Any help would be appreciated.
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"Luck is what happens when preparation meets opportunity." - Darrel K Royal Last edited by The_Dude; 09-10-2003 at 03:40 PM.. |
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#2 (permalink) |
Crazy
Location: Salt Lake City
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I stared at it for a full minute before I realized it wasn't a 5-part problem, it's multiple choice. The answer is #1. Just plug f(x) into that first equation (which is just the definition of the derivative) and you'll get a answer yourself easy enough.
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#5 (permalink) |
Upright
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Hey, let's see if I can remember how to do it, the answer that was previously mentioned (#1) is correct, and it can easily be solved by the chain rule[which is second nature, once you learn it]. However, what you are given here is the formal definition of a limit. To see how to get the answer of 6x+4 here are the complete mathematical steps:
Ok the first part says f(x+h), so that means take your f(x) equation and h to every x term you see: 3(x+h)^2 + 4(x+h) + 2 and that's basically it, now apply the equation lim as h->0 f(x+h) - f(x) / h so... we have: lim h->0 (3(x+h)^2 + 4(x+h) + 2 - [3x^2 + 4x + 2]) / h lim h->0 (3x^2 + 6xh + 3h^2 + 4x + 4h + 2 - [3x^2 + 4x + 2]) / h now, perform some cancellations and you have: lim h->0 (6xh + 3h^2 + 4h) / h now.. you see a common factor of h, which can also be cancelled, so you're left with: lim h->0 (6x+3h+4) ok, you may be saying what!?! that's not the answer.. now, you simply plug in 0 for h, so you have as a final answer: f`=dy/dx = 6x + 4 Hope this clarifies your problem and closes the book on this thread :-) P.S. Make sure you continuously write lim h->0 for each step as I did, teachers are notorious for taking points off for not including that, since it is a formal definition of a limit. |
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#6 (permalink) | |
Addict
Location: The Land Down Under
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Quote:
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Strewth |
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#10 (permalink) |
Know Where!
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http://mathworld.wolfram.com/Derivative.html
THIS IS THE SHORT WAY (sorta) 3x^2 + 4x + 2 ax^n + bx + c --- a,b,c, are constants (n*a)X^(n-1) +b --- c is a constant so it becomes 0 (3*2)X^(2-1) +4 =6X + 4 |
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#12 (permalink) |
Insane
Location: The Internet
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Some very good answers here.
Change f(x)= 3x^2 + 4x + 2 into "box" Then plug "box" into the differentiation formula in place of "x". you get: (box - h) - box / h now simplify the terms using the real value of "x" (ie. 3x^2 + 4x + 2) Later on, you will learn "the power rule" which simply states: where f(x)= x ^ 3 f ' (x)= [exp] x ^ [exp - 1] ie. f ' (x) = 3x^2 .. much better ![]() In your case: f(x)=3x^2 + 4x + 2 f ' (x)=6x + 4x + 0 PS: you will notice that the derivitave of any "real number" is always 0
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#14 (permalink) |
Insane
Location: The Internet
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Lol. It's not that bad ... really ...
![]() Honestly, I used to fear math .. until I decided - fuck it! I will work my ass off and actually master the topic. Incidently, I no longer have an ass .. but I do understand calculus. ![]()
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#15 (permalink) |
Addict
Location: The Land Down Under
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uh...how much of it, Sapper? Multivariate? Manifolds? If so, there's a senior lecturer around here who would love for you to explain it to him!
There's a lot more to Calc than derivatives and the kind of integrals you encounter in grade school!
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Strewth |
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#16 (permalink) |
Insane
Location: The Internet
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TIO, a tad anal or what? Wow.
You obviously can not appreciate the notion of understanding something. Not that it really matters - what you think is of no consequence to me. PS: Because you seem to lack fundamental social skills, I will take this to mean that you also lack fundemental logic skills. You should take special note that I said: "I will work my ass off and actually master the topic" and not "I will work my ass off and actually have mastered the topic".
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#17 (permalink) |
Invisible
Location: tentative, at best
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Heh heh - wait till 3 semesters from now, when your Differential Equations professor informs you that up until now, you've been doing it all backwards.
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