Proof: Equation x = a + bsin(x) has at Least One Solution

Proving the Equation x = a + bsin(x) Has a Solution

This proof demonstrates that the equation x = a + bsin(x), where a and b are positive constants, has at least one solution within the interval (0, a+b] using the Intermediate Value Theorem.

1. Defining the Function:

Let's define a function f(x) = x - a - bsin(x). Our goal is to prove there exists a value 'x' in the interval (0, a+b] such that f(x) = 0.

2. Applying the Intermediate Value Theorem:

The Intermediate Value Theorem states that if a continuous function f(x) changes sign over an interval, then it must equal zero at some point within that interval. We can apply this theorem to our function:

• Continuity: f(x) is continuous as it's a combination of continuous functions (linear and trigonometric).* Sign Change: * f(0) = 0 - a - bsin(0) = -a, which is negative. * f(a+b] = (a+b] - a - bsin(a+b]). Since -1 ≤ sin(x) ≤ 1, we know f(a+b] ≥ (a+b] - a - b = b, which is positive.

Since f(0) is negative and f(a+b] is positive, f(x) changes sign over the interval (0, a+b].

3. Conclusion:

By the Intermediate Value Theorem, because f(x) is continuous and changes sign over the interval (0, a+b], there must exist at least one value 'x' within this interval where f(x) = 0. This proves that the equation x = a + bsin(x) has at least one solution in the interval (0, a+b].