| Blogs > Raithed |
|
Raithed
China7078 Posts
f(x) = 9(x²-3)/x³ Of course, then we do the division rule... f`(x) = x³(18x)-(9)(x²-3)(3x²) = 9(9-x²)/x^4 Now it says "at the point (3,2), the value of the derivative is f`(3) = 0." I don't see how it gets 3,2? Can anyone figure it out? Doesn't make sense to me. The professor teaches wayyyyy too fast, someone be my mentor for this? I'm having a little problem with the quotient rule as well(maybe that's collapsing the rest). It's (bottom)*(d.top) - (top)*(d.bottom) / (bottom)^2 right?  | ||
|
fanatacist
10319 Posts
f(3) = 9(9-3)/27 = 9(6)/27 = 54/27 = 2 f(3) = 2 .: (3,2) f`(x) = 9(9-x²)/x^4 f`(3) = 9(9-9)/(3^4) = 0 .: Slope = 0 QED - At the X value 3, the Y value is 2 and the slope is 0. (3,2) m=0 Your quotient rule sounds right as well, if I remember correctly. | ||
|
boudiou
France190 Posts
Yes right The calculus is right here. df/dx is good so df/dx(3)=0 and if u check the value of your function at that point : f(3) =2 So on the graph of your function your min and or max (I didn't checked actually) is the point (3,2) because u determined (x,f(x)) after knowing x as a value nullifying the derivative. Going on historical stuff is a very good way to understand this subject. hope this helped I am not used to english terms in mathematics. :/ EDIT god damnit I was second and I am saying something wrong: df/dx doesn't necessarily means that there is either a maximum or a minimum. it is mainly used for that purpose though. | ||
|
Raithed
China7078 Posts
On October 28 2007 05:18 fanatacist wrote: f(x) = 9(x²-3)/x³ f(3) = 9(9-3)/27 = 9(6)/27 = 54/27 = 2 f(3) = 2 .: (3,2) f`(x) = 9(9-x²)/x^4 f`(3) = 9(9-9)/(3^4) = 0 .: Slope = 0 QED - At the X value 3, the Y value is 2 and the slope is 0. (3,2) m=0 Your quotient rule sounds right as well, if I remember correctly. Wow, this helped me so much. Thank you, I keep having the problem(s) with remembering my formulas because the prof. don't let us use it. But my question now is, I know where you are coming from with it, but you just stuck a x=3 in there just because or how did you figure that out? | ||
|
fanatacist
10319 Posts
On October 28 2007 05:38 Raithed wrote: Wow, this helped me so much. Thank you, I keep having the problem(s) with remembering my formulas because the prof. don't let us use it. But my question now is, I know where you are coming from with it, but you just stuck a x=3 in there just because or how did you figure that out? Glad to hear it. What do you mean won't let you use them? That's retarded, they are there for a reason - unless you are going over the foundation/explanations for the formulas right now, you should be able to use them. I got the number from you, because you said "(3,2)" something something, but were confused about the proof. So I showed you why it comes out to (3,2). Normally they way to find this X and vertex would be to set the f`(x) = 0 and find a solution from that, because the peak of a curve always has a slope of 0. So, I guess to answer your question by example, this would have been the process. f(x) = 9(x²-3)/x³ f`(x) = 9(9-x²)/x^4 9(9-x²)/x^4 = 0 [ 9(9-x²)/x^4 ] * (x^4)/9 = 0 * (x^4)/9 **CAREFUL!: This step cannot always be performed depending on the limits of the problem, such as the domain. If you do this you have to be sure you can, and if you get a messed up answer, it means you can't. In other cases, the answers you get out of this will not be the only possible answers. 9-x² = 0 .: x1 = -3, x2 = 3 (Possible vertices) f(x) = 9(x²-3)/x³ f(x1) = 9(x1²-3)/x1³ f(-3) = 9(9-3)/-27 f(-3) = 54/-27 = -2 .: (-3,-2) f(x2) = 9(x2²-3)/x2³ f(3) = 9(9-3)/27 f(3) = 54/27 = 2 .: (-3, -2), (3,2) are vertices. If you check the graph on a graphing calculator, you get those 2 points as obvious vertices, and with a vertical asymptote in the domain at X=0 (because you can't divide by 0). So, these solutions are correct and complete. That's how you would go about getting them. | ||
|
Raithed
China7078 Posts
I understand the derivatives and all that even though test taking isn't my best. 9(9-x²)/x^4 = 0 I get that. But when you did this: [ 9(9-x²)/x^4 ] * (x^4)/9= 0 * (x^4)/9 This really confused me, can you elaborate on it more, explain it in dumb terms so I can understand? It looks like quotient rule again but I don't want to take it into assumption. | ||
|
fanatacist
10319 Posts
Let me try to explain a bit more based on what you do get, then. f`(x) = 0 <-- How we get the local extrema 9(9-x²)/x^4 = 0 <-- f`(x) = 0 Now, what we have to do from this is get an expression where we can determine what specific x values make f`(x) = 0. So, we need to simplify. First thing to be noticed is that in the 9(9-x²), the 9 is irrelevant - what I mean by this is we can divide both sides by 9. Therefore, we get the first step of simplification: [9(9-x²)/x^4] /9 = 0 /9 <-- Dividing both sides by 9 to simplify We want the simplest expression possible. A way to prove that this doesn't affect the outcome is as follows: Let's take an arbitrary function and make it equal to zero. f(x) = x - 3 x - 3 = 0 .: x = 3 9(x - 3) = 9(0) 9(3 - 3) = 0 9(0) = 0 0 = 0 .: True. So what does this mean in our case? It proves the following: If f(x) = z, then c * f(x) = c * z <-- For any constant c f(x)/c = z/c <-- For any constant c This means we can remove the 9 from the equation to get: [9(9-x²)/x^4] /9 = 0 /9 <-- Dividing both sides by 9 to simplify (9-x²)/x^4 = 0 <-- Result The same proof can be done for a function of x, what we'll call g(x) g(x) * f(x) = g(x) * z <-- For any x in the domain. f(x)/g(x) = z/g(x) <-- For any x in the domain *NOTE: This is what I meant about being careful. Note how I said any x in the domain. That means that if x can't be equal to a certain constant, this may not work. For example: x²/x^4 = 0 <-- x^4 cannot be equal to 0 for this to work, because you cannot divide by zero, so we keep that in mind. (x²/x^4) * x^4 = 0 * x^4 <-- Multiplying both sides by x^4 x² = 0 x = 0 But, remember how we said x^4 cannot equal zero? Let us check: x = 0 x^4 = 0^4 x^4 = 0 This is not possible, because x^4 cannot equal zero! This is a case where you cannot simply divide by a function to simplify the equation. Be wary of these. Now, back to our nicer problem: (9-x²)/x^4 = 0 [(9-x²)/x^4] * x^4 = 0 * x^4 <-- Multiplying both sides by x^4 9-x^2 = 0 <-- Result x^2 = 9 x1 = -3 x2 = 3 Notice how neither are 0 (which would make x^4 = 0 and thus dividing by 0)... This means that these answers are legal according to our domain. So, we can divide by 9 and multiply by x^4 safely in this case to get the simplified equation. What the following operation does: [ 9(9-x²)/x^4 ] * (x^4)/9= 0 * (x^4)/9 Is combine both of those at the same time - it's not a derivative rule or anything. It's just multiplying by x^4 and dividing by 9 simultaneously. Wow, this came out super long, but if you really understand a concept, you will never make a mistake in it, right? n_n EDIT: Fixed a division to a multiplication somewhere up in there... Lols. | ||
|
Raithed
China7078 Posts
+ Show Spoiler + ![[image loading]](http://i24.tinypic.com/2a5deop.jpg) I get that much, sorry I am a dumbass. | ||
|
fanatacist
10319 Posts
![[image loading]](http://i74.photobucket.com/albums/i265/fanatacist-/mat.png) | ||
|
Raithed
China7078 Posts
| ||
|
fanatacist
10319 Posts
On October 28 2007 08:15 Raithed wrote: WOW, hahaha, holy shit. That... seemed, way easier than reading it. You're my hero fanatacist. Thanks for helping a fool like me. <3 Back to studying. ^____^ I see xD I was just trying to prove it in case you were like "How come you can do that?" or something. No problem <3 | ||
|
Meta
United States6225 Posts
On October 28 2007 05:06 Raithed wrote: This is not sinking in, I'm studying this section from the book, and one example goes as: f(x) = 9(x²-3)/x³ Of course, then we do the division rule... f`(x) = x³(18x)-(9)(x²-3)(3x²) = 9(9-x²)/x^4 Now it says "at the point (3,2), the value of the derivative is f`(3) = 0." I don't see how it gets 3,2? Can anyone figure it out? Doesn't make sense to me. The professor teaches wayyyyy too fast, someone be my mentor for this? I'm having a little problem with the quotient rule as well(maybe that's collapsing the rest). It's (bottom)*(d.top) - (top)*(d.bottom) / (bottom)^2 right? think of it this way: low d-hi minus hi d-low over low low. speak it out, "low dee-high minus high dee-low over low low" my calc 1 teacher taught me that and it really has stuck lol | ||
|
Raithed
China7078 Posts
On October 30 2007 01:31 Meta wrote: think of it this way: low d-hi minus hi d-low over low low. speak it out, "low dee-high minus high dee-low over low low" my calc 1 teacher taught me that and it really has stuck lol This lady told me it, she said it's a song, haha. | ||
|
JeeJee
Canada5652 Posts
On October 30 2007 04:04 Raithed wrote: This lady told me it, she said it's a song, haha. low low low d-high minus hi d-low. all over bottom squared and away we go to the tune of row row row your boat lols | ||
Chill
Calgary25990 Posts
Do you actually remember it like that?  | ||
|
fanatacist
10319 Posts
On October 31 2007 04:35 Chill wrote: roflroflrofl Do you actually remember it like that? People in America tend to be very special ): I prefer the logical reasons... If you can prove it to yourself, you will never forget it. Fuck songs. | ||
| ||
WardiTV 2025
Playoffs
Reynor vs MaxPaxLIVE!
SHIN vs TBD
TBD vs Cure
Solar vs herO
Classic vs TBD
TBD vs Clem
[ Submit Event ] |
|
StarCraft 2 StarCraft: Brood War Horang2 Dota 2 1601Stork 439Leta  399firebathero 373EffOrt  309Last 209Zeus 201Larva 194ggaemo 105Mong 52[ Show more ] Counter-Strike Heroes of the Storm Other Games Organizations Other Games StarCraft: Brood War StarCraft 2 StarCraft: Brood War |
|
Ladder Legends
BSL 21
Sziky vs Dewalt
eOnzErG vs Cross
Sparkling Tuna Cup
Ladder Legends
BSL 21
StRyKeR vs TBD
Bonyth vs TBD
Replay Cast
Wardi Open
Monday Night Weeklies
WardiTV Invitational
Replay Cast
[ Show More ] |
|
|
EPT
