curve intersection
- Thread starter mufc802
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if they intersect, then x and y must have the same values at some points on both curves. So y = y and x = x as it were.
You could write y1 = x(x-4), y2 = -1/x and then say that if the two points intersect at some x value then the two curves must occupy the same point, so y1 = y2. I dunno if that's an easier way to think about it or not..
So you can just write x(x-4) = -1/x and solve that thang. It'll end up as a cubic(!) here so you know there are between 0 and 3(!) intersection points, and the number of distinct solutions in the real numbers that you can find will basically tell you what those intersection points are on the x axis
Just subsitute them back into one of your original equations and happy days!
EDIT: I can't count!
You could write y1 = x(x-4), y2 = -1/x and then say that if the two points intersect at some x value then the two curves must occupy the same point, so y1 = y2. I dunno if that's an easier way to think about it or not..
So you can just write x(x-4) = -1/x and solve that thang. It'll end up as a cubic(!) here so you know there are between 0 and 3(!) intersection points, and the number of distinct solutions in the real numbers that you can find will basically tell you what those intersection points are on the x axis
Just subsitute them back into one of your original equations and happy days!
EDIT: I can't count!
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Substitute y=-1/x into the other equation to give -1/x = x(x-4)
multiply both sides by -x to give 1 = -x²(x-4)
rearrange to get 0 = x³ - 4x² + 1
solve the cubic to give you your x co-ordinates of the intersections. (in this case the answers are all real and distinct, but you could get repeated solutions at a tangential intersection or pairs of imaginary roots)
substitute these back into y=-1/x to give the corresponding y co-ordinates
EDIT: Its not an easy cubic to solve, the x values come out as 3.93543...., 0.53740... and -0.47283...
You'll have to use an iterative method to converge on the actual values if you are doing it by hand, or use a calculator with that function.
multiply both sides by -x to give 1 = -x²(x-4)
rearrange to get 0 = x³ - 4x² + 1
solve the cubic to give you your x co-ordinates of the intersections. (in this case the answers are all real and distinct, but you could get repeated solutions at a tangential intersection or pairs of imaginary roots)
substitute these back into y=-1/x to give the corresponding y co-ordinates
EDIT: Its not an easy cubic to solve, the x values come out as 3.93543...., 0.53740... and -0.47283...
You'll have to use an iterative method to converge on the actual values if you are doing it by hand, or use a calculator with that function.
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No idea what that means tbh, but if you're happy with solving cubics then basically what vertica said. Otherwise..Need something a bit more clever..
When the curves intersect, the y coordinates are the same (as are the x coordinates for that matter), so x(x-4) is equal to -1/x.
x(x-4) = -1/x
x^2(x-4) = -1
x^3 - 4x^2 + 1 = 0
Now, we can see from your graph that there is a root between x=2 and x=4. What we need to do is narrow the range until we are accurate to a certain amount. The expression we found earlier (x^3 - 4x^2 + 1) will flip its sign on either side of a root, i.e.
x = 2
y = 2^3 - 4*2^2 + 1 = -7
x = y
y = 4^3 - 4*4^2 + 1 = 1
Between 2 and 4, the sign has changed from negative to positive, so there's a root in that range.
Let's try x = 3.
y = 3^3 - 4*3^2 + 1 = -8
OK, so there's no sign change between 2 and 3, but there is between 3 and 4. That's halved the range to look within. The next value to try is 3.5.
y = -5.125
So the range is now from 3.5 to 4. Let's keep going
x = 3.75
y = -2.51.... (So x is between 3.75 and 4)
x = 3.875
y = -0.87... (So x is between 3.875 and 4)
x = 3.9375
y = 0.03... (So x is between 3.875 and 3.9375)
x = 3.90625
y = -0.43... (So x is between 3.90625 and 3.9375)
x = 3.921875
y = -0.20... (So x is between 3.921875 and 3.9375)
x = 3.9296875
y = -0.08... (So x is between 3.9296875 and 3.9375)
Now, the range is sufficiently tight enough to say that x = 3.93 to 2 d.p. That's one root found! Now for the other two. But I'll let you try those!
x(x-4) = -1/x
x^2(x-4) = -1
x^3 - 4x^2 + 1 = 0
Now, we can see from your graph that there is a root between x=2 and x=4. What we need to do is narrow the range until we are accurate to a certain amount. The expression we found earlier (x^3 - 4x^2 + 1) will flip its sign on either side of a root, i.e.
x = 2
y = 2^3 - 4*2^2 + 1 = -7
x = y
y = 4^3 - 4*4^2 + 1 = 1
Between 2 and 4, the sign has changed from negative to positive, so there's a root in that range.
Let's try x = 3.
y = 3^3 - 4*3^2 + 1 = -8
OK, so there's no sign change between 2 and 3, but there is between 3 and 4. That's halved the range to look within. The next value to try is 3.5.
y = -5.125
So the range is now from 3.5 to 4. Let's keep going
x = 3.75
y = -2.51.... (So x is between 3.75 and 4)
x = 3.875
y = -0.87... (So x is between 3.875 and 4)
x = 3.9375
y = 0.03... (So x is between 3.875 and 3.9375)
x = 3.90625
y = -0.43... (So x is between 3.90625 and 3.9375)
x = 3.921875
y = -0.20... (So x is between 3.921875 and 3.9375)
x = 3.9296875
y = -0.08... (So x is between 3.9296875 and 3.9375)
Now, the range is sufficiently tight enough to say that x = 3.93 to 2 d.p. That's one root found! Now for the other two. But I'll let you try those!
Yea bracketing algorithm or Newton will do it pretty easily, but I wonder if there's an easy analytical trick we're missing
EDIT: What level's this for btw? If it's uni or similar then man up and solve the cubic
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Now, the range is sufficiently tight enough to say that x = 3.93 to 2 d.p. That's one root found! Now for the other two. But I'll let you try those!
You should really close it in a bit further for 2d.p. - it is 3.94 to 2d.p.
/pedant
It's not about finding the actual co-ordinates of the intersection for these curves, it's just knowing that they actually intersect or not. The textbook specifically says "You would not be expected to solve this equation in C1".
This is the answer that "explains" what to do:
This is the answer that "explains" what to do:
Yea bracketing algorithm or Newton will do it pretty easily, but I wonder if there's an easy analytical trick we're missing
There are a few tricks, but I guessed that they are well beyond the level of this question.
I figured that interval bisection was the method of choice here.
You should really close it in a bit further for 2d.p. - it is 3.94 to 2d.p.
/pedant
Aw, crap
That'll teach me to rush these things!
Yes, it's 3.94 to 2 d.p.
It's not about finding the actual co-ordinates of the intersection for these curves, it's just knowing that they actually intersect or not. The textbook specifically says "You would not be expected to solve this equation in C1".
This is the answer that "explains" what to do:
LOL
ok, my bad, didnt realise what the question was getting at. Ok, what they're asking for is for you to figure out how many points of intersection there are and what quadrants the intersections lie in.So, they do something similar to what Morbius described and look at some points where one function is bigger than the other, and try to look for a point where it's less than. We know that x(x-4) is 0 at x = 0. Furthermore, -1/x < 0 for all x > 0. So can we find a point of x(x-4) such that x(x-4) < -1/x?
If so, then we must have 2 intersections in the 4th quadrant! The A-Level reason for this is that in order to get from the positive side of -1/x to the negative side and back again, we have to cut it twice! So what they do is say, well let's try x = 2. They find that in fact for x = 2, x(x-4) < -1/x, so there are 2 points of intersection in the 4th quadrant!
Then for the third one you can just sort of use the same principle. At x = -0.000000001 (note for the pedants that i didnt say 0), x(x-4) < -1/x (check!). So just find a point where it's greater than -1/x and happy days are had (e.g. x = -12 or something). So we have 3 intersections, with our last one in the 2nd quadrant
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There are a few tricks, but I guessed that they are well beyond the level of this question.
I figured that interval bisection was the method of choice here.
I dunno, couldve been a first year analysis question
Not that impressed with the textbook method.
"y = x(x-4) is a U shaped curve"
That's very nice, but it's hardly mathematical now is it?
And the whole method rests on this statement. Sigh.
It's A-Level
. In fairness you don't learn enough to be able to say any more until like..C2?Ah in which case its the part of C1 that requires you to have knowledge of functions and their shapes (although y=-1/x is not classed as a function IIRC due the discontinuity at x=0) the answer is fairly comprehensive, but i'll try to expand:
y=x(x-4) is a standard positive quadratic and you should therefore expect it to be a U shape.
Finding (if and) where it crosses the x axis by substituting y=0 into it, tells you that the bottom of the curve crosses the x axis at x=0 and x=4 and hence the lowest point its at x=2 (standard quadratic curve is symmetric).
The -1/x curve is again a standard shape you should know, its asymptotic to the axes in the second and fourth quadrants.
The curves will obviously cross in the second quadrant (upper left) as one is tending towards the y axis and the other away from it in the -x direction.
In the fourth quadrant, they have chosen a value (x=2) to see if the -1/x curve is above the quadratic at that point (compare y values) if it is then the two lines must have crossed. given that the quadratic goes back into the first quadrant a with +x, and the other line stays in the fourth quadrant they must cross again
based on this there are three intersects, who's locations have x values that are the solutions we've already derived above
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LOL
So, they do something similar to what Morbius described and look at some points where one function is bigger than the other, and try to look for a point where it's less than. We know that x(x-4) is 0 at x = 0. Furthermore, -1/x < 0 for all x > 0. So can we find a point of x(x-4) such that x(x-4) < -1/x?
If so, then we must have 2 intersections in the 4th quadrant! The A-Level reason for this is that in order to get from the positive side of -1/x to the negative side and back again, we have to cut it twice! So what they do is say, well let's try x = 2. They find that in fact for x = 2, x(x-4) < -1/x, so there are 2 points of intersection in the 4th quadrant!
Then for the third one you can just sort of use the same principle. At x = -0.000000001 (note for the pedants that i didnt say 0), x(x-4) < -1/x (check!). So just find a point where it's greater than -1/x and happy days are had (e.g. x = -12 or something). So we have 3 intersections, with our last one in the 2nd quadrant
So they used 2 because it's the mid-section of 0 and 4 so that would be the lowest point on the curve? If that meant that x(x-4) was less than -1/x it would mean it did intersect because it went lower on the y axis? If x(x-4) was greater than -1/x when x was 2 it would mean that it didn't go down enough on the y-axis to cut through -1/x?
What if they just touched rather than cut so would it be possible for just one intersection?
All that it tells us is that it was greater (at x = 0), then it was lower (at x = 2). Then at x = 4 it became greater again. Now, this must mean that we passed through at least two intersections. First to get lower, then again to become greater again.
If it just touched and went back up again, then it wouldnt necessarily be lower. So if we actually found that it was greater at x = 2, then we couldnt say anything about the intersections, we'd need to look at more points.
You could differentiate y = x^2 - 4x to get dy/dx = 2x - 4
y(x) achieves its critical point at dy/dx = 0, which in this case is x = 2, y = -4
We know this point is a minimum as d2y/dx2 = 2 which is greater than 0
At x = 2, y = -1/2 for the second curve. i.e. the only minimum of y = x(x-4) falls below the curve y = -1/x, so there is two intersections in that quadrant.
Now take say x = -2, then y = x(x-4) = 12 while y = -1/x = 1/2 So x(x-4) is greater than -1/x at x = -2
Now take x = -0.1, then y = x(x-4) = 0.41 while y = -1/x = 10 So x(x-4) is less than -1/x at x = -0.1
As both curves are continuous on [-2,-0.1], they must intersect.
So three distinct points of intersection.
Though solving the cubic x(x-4) = -1/x is probably easier
y(x) achieves its critical point at dy/dx = 0, which in this case is x = 2, y = -4
We know this point is a minimum as d2y/dx2 = 2 which is greater than 0
At x = 2, y = -1/2 for the second curve. i.e. the only minimum of y = x(x-4) falls below the curve y = -1/x, so there is two intersections in that quadrant.
Now take say x = -2, then y = x(x-4) = 12 while y = -1/x = 1/2 So x(x-4) is greater than -1/x at x = -2
Now take x = -0.1, then y = x(x-4) = 0.41 while y = -1/x = 10 So x(x-4) is less than -1/x at x = -0.1
As both curves are continuous on [-2,-0.1], they must intersect.
So three distinct points of intersection.
Though solving the cubic x(x-4) = -1/x is probably easier