
Gregg O. answered 12/27/15
Tutor
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Cal Poly Pomona engineering valedictorian, expert in geometry
The three conditions for Rolle's Theorem to hold on an interval [a,b] are
(i) f(x) is continuous on [a,b]
(ii) f(x) is differentiable on (a,b)
(iii) f(a) = f(b)
So, let's prove that the function satisfies these conditions.
(i) cos (2x) is continuous on (-∞, ∞), so it is also continuous on [pi/8, 7*pi/8].
(ii) The derivative of f(x) is -2sin(2x), which exists on [pi/8, 7*pi/8].
(iii) f(pi/8) = cos (pi/4) = √(2)/2
f(7*pi/8) = cos (7*pi/4) = √(2)/2.
So, the function satisfies the 3 hypothesis.
Next, we are guaranteed that there is at least one number c∈(pi/8, 7*pi/8) such that f'(c) = 0. So, let's solve:
f'(x) = -2sin(2x)
-2sin(2x) = 0
sin(2x) = 0
A variable substitution may be helpful:
Let u = 2x. We are given that pi/8 < x < 7*pi/8, which means that
pi/4 < u < 7*pi/4.
On this interval, sin(u) = 0 has the single solution u = pi, which means
2x = pi
x = pi/2.
So, there is one c in (pi/8, 7*pi/8) which satisfies the theorem, which is c = pi/2.
Mike H.
f'(x) = -2sin(2x) is the derivative.. -2sin(2x) = 0 to satisfy the condition f`(c) = 0... Solve for sin: divide both sides by -2 and you get "sin(2x) = 0"
Report
07/11/22
Micahya H.
f'(x) = -2sin(2x) -2sin(2x) = 0 sin(2x) = 0 Can you explain what happened here10/13/21