Let f(x) = 2x + 1 − sin(x), how many roots does f(x) have in the interval [−π, π]? Use the next steps to prove that it has only one root.
a) Use the Intermediate Value Theorem to show that f(x) has at least one root.
(b) Explain why f(x) is increasing on the interval [−π,π].
(c) Use the Mean Value Theorem or Rolle’s Theorem, along with your answer from part (b), to show that f(x) cannot have two roots in the interval [−π, π]. (Hint: If there are two roots, what might go wrong?)
a peek at the graph might guide you along
http://www.wolframalpha.com/input/?i=plot+y+%3D++2x+%2B+1+%E2%88%92+sin%28x%29
a) To use the Intermediate Value Theorem, we need to show that f(x) takes both positive and negative values in the interval [-π, π].
Let's evaluate f(x) at the endpoints of the interval:
For x = -π,
f(-π) = 2(-π) + 1 - sin(-π) = -2π + 1
For x = π,
f(π) = 2(π) + 1 - sin(π) = 2π + 1
Since -2π + 1 is negative and 2π + 1 is positive, f(x) takes both positive and negative values at the endpoints of the interval.
By the Intermediate Value Theorem, since f(x) is continuous in the interval [-π, π] and it takes both positive and negative values, it must have at least one root within that interval.
b) To prove that f(x) is increasing on the interval [-π, π], we need to show that the derivative, f'(x), is greater than or equal to zero for all x in that interval.
Let's find the derivative of f(x):
f'(x) = d/dx (2x + 1 - sin(x))
= 2 - cos(x)
To prove that f(x) is increasing, we need to show that f'(x) ≥ 0 for all x in the interval [-π, π].
Using the fact that cos(x) ranges between -1 and 1 for any x, we can see that 2 - cos(x) will always be greater than or equal to zero. Therefore, f(x) is increasing on the interval [-π, π].
c) Assume f(x) has two roots within the interval [-π, π]. Let's call these roots x_1 and x_2, where x_1 < x_2.
By Rolle's Theorem, if f(x) has two roots, then there must be at least one point c in the interval (x_1, x_2) where f'(c) = 0.
However, we have previously shown that f(x) is increasing on the interval [-π, π]. This means that f'(x) > 0 for all x in that interval, and there cannot be any value of c where f'(c) = 0.
Therefore, f(x) cannot have two roots within the interval [-π, π].
Hence, we have proven that f(x) has only one root in the interval [-π, π].