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Stuck on basics: calc.
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Posted by: NightMage
Well I know this is really basic stuff but I always forget it... how do I get my zero's i know in an equation like
y=x^3 - 3x^2 + 4
you get one by putting replacing y for 0 but i can't remember if with the other one you replac x for 0.... I did that once and was told that zero was wrong.
and I know most of you guys are doing calc 
Posted by: uh...ok
Yes, to find the zeros you set x = 0 to find y and set y = 0 to find x.
In this case, the easy one is when x = 0, y = 4. Therefore, one of your zeros is at point (0,4).
For the other point, you set y = 0, and solve for x.
So x^3 - 3x^2 + 4 = 0. To start you off, one of the solutions is (-2,0). 
EDIT: If you graph it, you'll find it's the ONLY solution.
-uh...ok
Posted by: NightMage
ok so you got it by using the derivative? because i worked out that it's the same as the stationary point... but i have (2,0)
though i can see how you may have gotten (-2,0) you just subtracted 3x ^2 instead of adding 6x
Posted by: redwench
-8 - 3(4) + 4 != 0 youhawk
it has to be a positive number for this particular equation :P
Posted by: NightMage
thank god so I wasn't going insane I swear I'm gonna try and avoid doing calc ever again lol
Posted by: Lord_Buttplug
The zeros that I see are at (-1,0) and (2,0).
Use this to solve, I think.
x^3-3x^2+4=0
x^3-3x^2=-4
x^2(x-3)=-4
x-3=-4 => x = -1
This is the part I can't remember.
x^2 = -4 You get the (2,0) intercept from it, but my brain is dead from no sleep for last few days.
Posted by: Gunslinger
The roots are:
-1
2 with a multiplicity of 2
That is the complete answer.
Posted by: NightMage
This is how I did it to get (2,0)
f(x) = x^3 - 3x^2 + 4
f ' (x) = 3x^2 - 6x
0 = 3x^2 - 6x
6x = 3x^2
2 = x
Posted by: uh...ok
That gives you the zero of f'(x), not f(x) buddy.
And yes, I did make a mistake... it's (2,0), not (1,0). And the other zero IS (0,4) (very easily proven if you just plug in 0 for x, then y=4. And LPB is also correct in finding the third zero (-1,0). I didn't notice that on my graph.
Coincidentally, two of the zeros of this function also happen to be the relative extrema of this function. Which explains why deriving the function gives you x=2 and x=0.
(You lost an x there, buddy, when you cancelled them out. You can't do that when looking for extrema.)
You were doing ok up to the point where you did:
3x^2 = 6x and cancelled.
You should have factored:
x(3x-6) = 0
Which gives you the solutions x = 0 and x = 2. Do the first derivative test and you'll find that f(x) has a relative maximum at point (0,4) and a relative minimum at (2,0).
I must stress once again, however, that the extrema fall on the zeros by pure coincidence. You do NOT find the zeros of a function by taking its derivative. You find the zeros of a function by setting x to zero and solving for y and setting y to zero and solving for x.
-uh...ok
Posted by: NightMage
hehe i lost the x because i divided both sides by 3x
first off i needed to get the stationary points.... i got the y intercept (maxima) and one of the x zeros (minima) I think my prob was going back to the original function and trying to get the zeros from that which just didn't want to work.
I kinda figured it was a fluke that my stationarys were the same as my zeros
Posted by: uh...ok
And now for the tedious lesson on finding zeros of a function that's higher order than a quadratic (where you can simply use the quadratic formula) and where you cannot factor out anything to make things simpler.
Say you have a function:
f(x) = x^3 - 3x^2 +4.
To find the points where f(x) = 0, however, is much more complicated. But it's doable - even by hand. But it's also very, very tedious. (For our purposes, pretend that we cannot simply just factor out x and solve).
What you do is look at the coefficient of the highest order variable and the constant. Let the constant be p and the coefficient of the highest order variable be q.
p = 4
q = 1
Now here comes the tricky part. What you want to do is find a constant c for (x - c), which is a factor of f(x). Basically you want to transform the function f(x) into something that's simpler. You want:
f(x) = (x - c) g(x)
g(x) = f(x)/(x - c)
What is c, then? The possible values of c are given by the factors of p divided by the factors of q.
Factors of p: -4, -2, -1, 1, 2, 4
Factors of q: 1, -1
c = p/q = -4, -2, -1, 1, 2, 4
So the possible values of c are -4, -2, -1, 1, 2, and 4.
Now which one is the right factor? You test them. Each and every possible value of c. The way to test them? Divide f(x) by (x-c). Yes, long division of functions, lots of fun. You basically divide until you find a c that has no remainder.
Now since we already figured this problem out the easy way, there's no point in going through all the tests and I'll just pick a number that we know works. Since we know that a zero exists on x = 2, that means that one factor of this equation is at (x - 2), or c = 2.
Now I'm not sure if you learned to do long division this way, but it's a lot easier than actually using the long division method.
c=2|
1 _-3 _0 _4 (ignore the _s on this line, I used them as space fillers)
__ 2 -2 -4 (ignore the __ on this line also)
___________
1 -1 -2 0
Basically what you do is set up a line of numbers for the coefficients from highest to lowest order. Then you add the first column (1 + 0), multiply that sum by c (1 x 2), and that becomes the value for the second row on the second column, which is two. Repeat this until you get to the end. The zero on the last column means that you have a remainder of zero.
Which now leaves you with 1 -1 -2 0.
What do these numbers mean? Well when you do division of a power by (x-c) what happens is the function goes down by one exponent. The quotient of the division is the new function you get (or g(x)). In this case, these numbers represent the coefficients of your function. So the function f(x) can now be rewritten as the product of the factors, (x - c) and g(x).
f(x) = (x - 2) (x^2 - x - 2)
Factor the second factor and you get:
f(x) = (x - 2) (x + 1) (x - 2)
or
f(x) = (x+1) * (x - 2)^2
Which means f(x) = 0 at x = -1 and at x = 2.
Plug those values for x back into f(x), and you get your y-value for those points, which, obviously, is 0.
So in conclusion, you have the two points:
(-1,0) and (2,0) as your zeros. But as Gunny points out in the following post, (2,0), which is a critical point that touches but does not cross the x-axis, has a multiplicity of 2 (note the (x - 2)^2 means that the answer is x = 2 twice).
-uh...ok
Posted by: Gunslinger
Uhawk, I hope he didn't turn in that answer, because it is extremely wrong.
You guys are making this way more difficult than it actually is.
It's a cubic function, which always has 3 roots. In this case, all of them are real roots.
These roots are: -1, 2, and 2 (yes, that is '2' twice). So, the solution points are (-1,0),(2,0), and (2,0) (yes, twice).
Hence why I said earlier:
-1
2 with a multiplicity of 2.
Uhawk, in all your higher math, you have forgotten your easy math. Inputting '0' in for X and solving does not give you a root, it gives you a y-intercept. A root is not a y-intercept, however, but an x-intercept. When you input '0' you get '4' thus the point (0,4) as you said. that isn't a root. It's the y-intercept. The graph does not cross the x-axis at 0. That should take care of that little confusion.
Secondly, the point (2,0) is a root twice because the graph intersects the x-axis at that point but does not cross the x-axis at that point. The curve touches, then goes back up. This is called multiplicity, or a double root.
Solutions: -1, 2, 2
End of problem.
Posted by: uh...ok
Gunny I wasn't making it more difficult than it was. As I pointed out in the post, I was saying that the method is used for higher order equations where you simply can't just solve for x.
And yes, it is a root twice, as the equation f(x) = (x - 1) (x - 2)^2 demonstrates.
And yea, it was my mistake to include the y-intercept... I have forgotten much basic math. You're right, the zeros, or roots of a function are only for f(x) = 0.
The method above still stands, however, as it does give you the roots. The solutions remain -1 and 2, though, although one might want to mention that x = 2 is a double root.
Sorry for the confusion. I'll go back up there and correct myself.
BTW Gunny, the method I showed above is not even close to higher math. I learned that during my freshman year of high school. I remember it so well because it's such a damn good method. Usually I just have trouble remembering how to do the long division the easy way. 
-uh...ok
Posted by: Gunslinger
By higher math, I think I was referring to the people trying to use derivatives and such. Don't remember if you said that or not.
And yeah, I remember the rational root theorem. I also learned that in 9th grade. I hated that theorem.
Posted by: uh...ok
No, I didn't use derivatives; I was explaining to NightMage why taking the derivative doesn't work in this case.
And that's what it's called! Rational Root Theorem! Couldn't remember its name for the life of me.
And I guess this topic "Stuck on basics: calc" doesn't work quite well because it's more like "Stuck on basics: algebra". This skill of finding zeros, however, is crucial in Calc because you'll need it in almost all the time... e.g. finding critical points, finding points of inflection.
-uh...ok
Posted by: NightMage
hehe ok you guys have lost me lol.... where are you guys from, cos where i am we don't learn anything like that in grade 9 lol
Posted by: uh...ok
NightMage, I'm originally from San Francisco and that was taught when I was a freshman at my HS there. It was in an Adv.Algebra2/Trigonometry class.
The Rational Roots Theorem is quite handy to know, although you won't need it when you get a TI-82 or higher graphing calculator, since those can graph functions and calculate roots for you.
-uh...ok
Posted by: Cheese
I'm assuming you're forgetting to tell him that Adv. Algebra/Triginometry as a freshman is 2 years ahead of the typical high school student. Because no high school that I know starts kids off in advanced algebra. From my experience, it was algebra, then geometry, then alg 2, and pre-cal.
There was a large group of people like myself that took algebra in jr. high and started a year ahead, and a smaller group of kids like you who began high school 2 years ahead.
Posted by: Erekose
yeah uhok just forgets hes a mathmeatical freak of nature at times... but hey i can count my toes so 
...what do you mean take my socks off and do it again? but then there is more than one
Posted by: uh...ok
Well I just figured that since he's in Calculus by now, he's obviously taken Adv Alg.2/Trig. That's all.
-uh...ok
Posted by: Gunslinger
Yeah. I also took a combined Algebra2/Trigonometry class my freshman year in St.Louis. A few other freshman were in the class, but it was mostly juniors.
Posted by: uh...ok
My class was all freshman.
-uh...ok
Posted by: NightMage
lol.... this is the prob.... in Tasmania/Australia we don't have adv. algebra/trig. we're very basic. Algebra I can moderately understand, about the most we learn in trig before getting to pre-calc is graphing tan, cos, and sin. and the more basic stuff. even in the course i'm doing now trig is basic
now why didn't we just go
0 = x^3 - 3x^2 + 4
therefore 0 = (x-2)(x-2)(x+1)
i like trial and error, can save a lot of time
Posted by: NightMage
ok here's something a little more difficult
if f(x)= 1/3x^3 + x^2 + x + 1, find
i) the tangent to the curve that is parallel to the x-axis.
the second part is just to find it on a 45deg slope (reasonably easy)
Posted by: uh...ok
NightMage, the Rational Roots Theorem is basically what's used to find those factors you named.
As for your problem,
f(x) = 1/3x^3 + x^2 + x + 1
f'(x) = x^2 + 2x + 1 = (x + 1)^2
The tangent to the curve that is parallel to the x-axis means the tangent to the curve where the slope is zero.
f'(x) = 0 at point (-1, 2/3),
Therefore the point tangent to the curve that is parallel to the x-axis is (-1, 2/3).
If it wants the line tangent to the curve, use the point-slope formula.
(y - y-sub-0) = m(x - x-sub-0)
(y - 2/3) = 0
y = 2/3
-uh...ok
Posted by: NightMage
yeah I worked it out not long after I posted, I feel like a dumb ass lol.... so glad I'm not gonna continue with calc after this yr
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