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