If 𝑓(𝑥) = 𝑠𝑖𝑛−1𝑥, 0 < 𝑎 < 𝑏 < 1, use Mean value theorem to prove that𝑏 − 𝑎√1 − 𝑎2 < 𝑠𝑖𝑛−1𝑏 − 𝑠𝑖𝑛−1𝑎 < 𝑏 − 𝑎√1 − 𝑏2

The Mean Value Theorem (MVT) states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that the derivative of the function at that point is equal to the average rate of change over the interval.

The derivative of the function f(x) = sin^(-1)x is 1/√(1 - x^2).

According to the MVT, there exists a c in (a, b) such that:

f'(c) = (f(b) - f(a)) / (b - a)

Substituting the given function and its derivative, we get:

1/√(1 - c^2) = (sin^(-1)b - sin^(-1)a) / (b - a)

Rearranging the terms, we get:

(b - a) / √(1 - c^2) = sin^(-1)b - sin^(-1)a

Since c is in the interval (a, b), and 0 < a < b < 1, we know that a < c < b. Therefore, 1 - b^2 < 1 - c^2 < 1 - a^2. Taking the square root of these inequalities and inverting them, we get:

√(1 - a^2) < √(1 - c^2) < √(1 - b^2)

Substituting this into our previous equation, we get:

b - a < (b - a) / √(1 - a^2) < sin^(-1)b - sin^(-1)a < (b - a) / √(1 - b^2)

Multiplying through by the denominators, we get the desired result:

b - a√(1 - a^2) < sin^(-1)b - sin^(-1)a < b - a√(1 - b^2)