using the mean valiue thereom on f(x)=arcsin(x),
show that x< arcsin(x)< x/(sqrt1-(x^2)) for 0<x<1
To apply the Mean Value Theorem on the function f(x) = arcsin(x) in the interval (0, 1), we first need to verify that this function:
1. Is continuous on the closed interval [0, 1],
2. Is differentiable on the open interval (0, 1).
Since f(x) = arcsin(x) is the inverse of the sine function, the domain of f(x) is [-1, 1]. However, for this problem, we are only concerned with the interval (0, 1), for which f(x) = arcsin(x) is continuous and differentiable.
Now, we can proceed to apply the Mean Value Theorem.
According to the Mean Value Theorem, if a function f(x) 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 open interval (a, b) such that:
f(b) - f(a) = f'(c) * (b - a)
In our case, let a = x and b = x/(sqrt(1 - x^2)). The function f(x) = arcsin(x) is continuous and differentiable on the interval (0, 1).
Applying the Mean Value Theorem:
arcsin(x/(sqrt(1 - x^2))) - arcsin(x) = f'(c) * [x/(sqrt(1 - x^2)) - x]
= f'(c) * [x/(sqrt(1 - x^2)) - (x * sqrt(1 - x^2))/(sqrt(1 - x^2))]
Now, we need to find the derivative of f(x) = arcsin(x).
The derivative of f(x) = arcsin(x) can be found using the chain rule. Let u = x, so arcsin(x) = arcsin(u).
Then, the derivative of f(x) = arcsin(x) is:
f'(x) = d/dx [arcsin(x)]
= d/du [arcsin(u)] * d/dx [u] (by the chain rule)
= 1/sqrt(1 - u^2) * 1 (since d/du [arcsin(u)] = 1/sqrt(1 - u^2))
Now we can substitute f'(c) back into the Mean Value Theorem equation:
[arcsin(x/(sqrt(1 - x^2))) - arcsin(x)] = [1/sqrt(1 - c^2)] * [x/(sqrt(1 - x^2)) - (x * sqrt(1 - x^2))/(sqrt(1 - x^2))]
Simplifying:
[arcsin(x/(sqrt(1 - x^2))) - arcsin(x)] = [1/sqrt(1 - c^2)] * x * (1 - sqrt(1 - x^2))
Now, we have obtained the equation using the Mean Value Theorem. To complete the proof, we need to show that x < arcsin(x) < x/(sqrt(1 - x^2)) for 0 < x < 1.
First, let's show that x < arcsin(x):
Since f(x) = arcsin(x) is defined for x in (0, 1), it is known that arcsin(x) is always positive in this interval. Furthermore, the graph of f(x) = arcsin(x) is increasing for x in (0, 1). Therefore, since x is positive and less than x/(sqrt(1 - x^2)), we can conclude that x < arcsin(x).
Next, let's show that arcsin(x) < x/(sqrt(1 - x^2)):
Rearranging the equation obtained from the Mean Value Theorem:
arcsin(x/(sqrt(1 - x^2))) = arcsin(x) + [1/sqrt(1 - c^2)] * x * (1 - sqrt(1 - x^2))
Since x is positive and c is in the interval (0, 1), [1/sqrt(1 - c^2)] is also positive. Therefore, the right-hand side of the equation is positive.
Since f(x) = arcsin(x) is increasing for x in (0, 1), and x/(sqrt(1 - x^2)) is also positive and greater than x, we can conclude that arcsin(x) < x/(sqrt(1 - x^2)).
Combining both inequalities, we have:
x < arcsin(x) < x/(sqrt(1 - x^2)) for 0 < x < 1.
This completes the proof using the Mean Value Theorem on f(x) = arcsin(x).