Inactive Tutor answered 03/12/21
At the end points of interval [0, 2] f(0) = 2cos0 = 2 = 2cos2π = f(2).
f(x) continuis on [0, 2] and differentiable on (0, 2).
So by Rolle's Theorem f'(c) = - 2πsinπc = 0; sinπc = 0; πc = π; c = 1
Oluwatobi L.
asked 03/12/21f(x) = 2 cos 𝜋x, [0, 2]
If Rolle's theorem can be applied, find all values of c in the open interval (a, b) such that f '(c) = 0.
(Enter your answers as a comma-separated list. If Rolle's Theorem cannot be applied, enter NA.)
c =
Inactive Tutor answered 03/12/21
At the end points of interval [0, 2] f(0) = 2cos0 = 2 = 2cos2π = f(2).
f(x) continuis on [0, 2] and differentiable on (0, 2).
So by Rolle's Theorem f'(c) = - 2πsinπc = 0; sinπc = 0; πc = π; c = 1
Get a free answer to a quick problem.
Most questions answered within 4 hours.
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.