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If ๐‘“(๐‘ฅ) = ๐‘ ๐‘–๐‘›โˆ’1๐‘ฅ, 0 < ๐‘Ž < ๐‘ < 1, use Mean value theorem to prove that๐‘ โˆ’ ๐‘Žโˆš1 โˆ’ ๐‘Ž2 < ๐‘ ๐‘–๐‘›โˆ’1๐‘ โˆ’ ๐‘ ๐‘–๐‘›โˆ’1๐‘Ž < ๐‘ โˆ’ ๐‘Žโˆš1 โˆ’ ๐‘2

Question

If ๐‘“(๐‘ฅ) = ๐‘ ๐‘–๐‘›^{-1}๐‘ฅ, 0 < ๐‘Ž < ๐‘ < 1, use Mean value theorem to prove that

(b - a \sqrt{1 - a^2} < \sin^{-1}b - \sin^{-1}a < b - a \sqrt{1 - b^2} <

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Solution

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

Rearranging the terms, we get:

(b - a) / โˆš(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:

โˆš(1 - a^2) < โˆš(1 - c^2) < โˆš(1 - b^2)

Substituting this into our previous equation, we get:

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

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

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

This problem has been solved

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