Expert: Paul Klarreich - 12/7/2011

Question
Prove that 2π+x+2πsinx = 0 has at least two solutions on [-π,0]. I'm a bit stumped.

Answer
Questioner: Granit
Country: Ontario, Canada
Category: Calculus
Private: No
Subject: Calc. Rolles theorem
Question: Prove that 2π+x+2πsinx = 0 has at least two solutions on [-π,0]. I'm a bit stumped.
.............................
Let f(x) = 2π + x + 2πsinx

f(-pi) = 2pi - pi + 2pi sin(-pi)

Sin(-pi) = 0, so:

f(-pi) = pi

f(0) = 2pi + 0 + 2pi sin(0)

sin(0) = 0, so:

f(0) = 2pi

.........................

So we have f(x) positive at the endpoints.

f(-pi/2) = 2pi - pi/2 + 2pi sin(-pi/2)

sin(-pi/2) = - 1,  so:

f(-pi/2) = 2pi - pi/2 - 2pi = - pi/2

...............................

Now the Intermediate Value Theorem says:

If
1. f(x) is continuous on [a,b], and
2. f(a) = A, f(b) = B, and C is any number between A and B,
then
we can find c in (a,b) where f(c) = C.

In other words, f(x) takes on every value between its endpoint values.

If f(-pi) is positive, and
f(-pi/2) is negative, then


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