Find an equation for the cosine function that has one maximum at the point (4,6) and its next minimum at the point (5.5,1).
y(x) = A*cos(k*x + (phi))
y'(x) = -A*k*sin(k*x + (phi))
y''(x) = -A*k^2*cos(k*x + (phi)) = -k^2*y(x)
So we know that (4, 6) is a maximum point. That is, we know that
y'(4) = -A*k*sin(4k + (phi)) = 0
y''(4) = -k^2*y(4) = -A*k^2*cos(4k + (phi)) < 0 (So y(4) > 0)
This means:
sin(4k + (phi)) = 0
4k + (phi) = n*(pi) <-- n is some integer.
We know that (11/2, 1) is a minimum. That is, we know that
y'(11/2) = -A*k*sin(11k/2 + (phi)) = 0
y''(11/2) = -k^2*y(11/2) = -A*k^2*cos(11k/2 + (phi)) > 0 (So y(11/2) < 0)
Thus
sin(11k/2 + (phi)) = 0
11k/2 + (phi) = m*(pi) <-- m is some integer.
So we have two equations in k and (phi):
4k + (phi) = n*(pi)
11k/2 + (phi) = m*(pi)
The solution is:
k = 2(m - n)(pi)/3
(phi) = (11n - 8m)(pi)/3
So
y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
Where
y(4) > 0 and y(11/2) < 0
There are many many possibilities for this. Simply by playing around with m and n I find that m = 2 and n = 1, with A = -1 gives a good function. That is, one solution is
y(x) = -cos((2(pi)/3)*x - (5(pi)/3))
-Dan
try topsquark's method using y(x) = A*cos(k*x + (phi)) + C as the general form of the cosine function. finding that vertical shift C should solve your problems
or just alter topsquark's formula. say his minimum was at (5.5, -1) then just add 2 to his formula. that would cause a vertical shift of 2 units up, making the min point (5.5,1), of course that depends on whether or not his max point would fall in the right place
Okay, let's do this again. Most of my previous solution still holds, it just doesn't finish the job.
The m = 2, n = 1 part still works because it gives the location of the x values where we have the max and mins, though in reality x = 4 is a minimum point and x = 11/2 is a maximum. (That's why I had a negative A in my first solution.)y(x) = A*cos(k*x + (phi))
y'(x) = -A*k*sin(k*x + (phi))
y''(x) = -A*k^2*cos(k*x + (phi)) = -k^2*y(x)
So we know that (4, 6) is a maximum point. That is, we know that
y'(4) = -A*k*sin(4k + (phi)) = 0
y''(4) = -k^2*y(4) = -A*k^2*cos(4k + (phi)) < 0 (So y(4) > 0)
This means:
sin(4k + (phi)) = 0
4k + (phi) = n*(pi) <-- n is some integer.
We know that (11/2, 1) is a minimum. That is, we know that
y'(11/2) = -A*k*sin(11k/2 + (phi)) = 0
y''(11/2) = -k^2*y(11/2) = -A*k^2*cos(11k/2 + (phi)) > 0 (So y(11/2) < 0)
Thus
sin(11k/2 + (phi)) = 0
11k/2 + (phi) = m*(pi) <-- m is some integer.
So we have two equations in k and (phi):
4k + (phi) = n*(pi)
11k/2 + (phi) = m*(pi)
The solution is:
k = 2(m - n)(pi)/3
(phi) = (11n - 8m)(pi)/3
So
y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
Where
y(4) > 0 and y(11/2) < 0
We need to modify my solution a bit so that we have
y(x) = A*cos(k*x + (phi)) + C
as Jhevon suggested.
Specifically, I'm going to work with:
y(x) = A*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi))
+ C
Now, the cosine function varies between -1 and 1. We need it to vary between 1 and 6. To do this we need to adjust C such that it is right in the middle of the range [1, 6], that is we need C to be the average of 1 and 6:
C = (1/2)*(1 + 6) = 7/2
Now we need to "stretch" the cosine function so that it will go below the line y = 7/2 down to 1 and above the line y = 7/2 up to 6. Thus we need |A| to be:
|A| = (7/2) - 1 = 6 - (7/2) = 5/2
As I said before I need A to be negative given my choice of m and n, so I get:
y(x) = -(5/2)*cos([2(m - n)/3](pi)*x + [(11n - 8m)/3](pi)) + (7/2)
-Dan