Answer :
dy/dx = 1 + cos(x).
To find all extrema in [0, 2π], we set dy/dx = 0 to obtain:
0 = 1 + cos(x)
==> cos(x) = -1.
This gives a solution of x = π on [0, 2π].
Therefore, the extrema on the intervals [0, 2π] is at x = π, which produces a point of (π, π).
I hope this helps!
θ βяια
By taking derivatives:
dy/dx = 1 + cos(x).
To find all extrema in [0, 2π], we set dy/dx = 0 to obtain:
0 = 1 + cos(x)
==> cos(x) = -1.
This gives a solution of x = π on [0, 2π].
Therefore, the extrema on the intervals [0, 2π] is at x = π, which produces a point of (π, π).
I hope this helps!
dy/dx = 1 + cos(x).
To find all extrema in [0, 2π], we set dy/dx = 0 to obtain:
0 = 1 + cos(x)
==> cos(x) = -1.
This gives a solution of x = π on [0, 2π].
Therefore, the extrema on the intervals [0, 2π] is at x = π, which produces a point of (π, π).
I hope this helps!