# Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem #f(x)=x-cosx#; [-pi/2, pi/2]?

See below.

I am unaware of anything that might be called "the principle of the MVT" that can be used to do this.

What we need is knowledge of what the conclusion of MVT says and enough algebra and trigonometry skills to solve an equation.

Solution

The conclusion of MVT for this function on this interval says:

This amounts to solving

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To find all numbers ( c ) that satisfy the conclusion of the Mean Value Theorem for the function ( f(x) = x - \cos(x) ) on the interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ), we need to follow these steps:

- Verify that the function ( f(x) ) is continuous on the closed interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ) and differentiable on the open interval ( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) ).
- Compute the derivative of the function ( f(x) ) with respect to ( x ), denoted as ( f'(x) ).
- Apply the Mean Value Theorem, which states that 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'(c) = \frac{f(b) - f(a)}{b - a} ).
- Substitute the values of ( a ), ( b ), ( f(a) ), and ( f(b) ) into the Mean Value Theorem formula.
- Solve for ( c ) to find the specific numbers that satisfy the conclusion of the theorem.

Given the interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ), ( a = -\frac{\pi}{2} ) and ( b = \frac{\pi}{2} ). Thus,

[ f(a) = f\left( -\frac{\pi}{2} \right) = -\frac{\pi}{2} - \cos\left( -\frac{\pi}{2} \right) = -\frac{\pi}{2} + 1 ]

and

[ f(b) = f\left( \frac{\pi}{2} \right) = \frac{\pi}{2} - \cos\left( \frac{\pi}{2} \right) = \frac{\pi}{2} - 1 ]

Now, compute the derivative ( f'(x) ) of the function ( f(x) ):

[ f'(x) = 1 + \sin(x) ]

Applying the Mean Value Theorem, we have:

[ f'(c) = \frac{f(b) - f(a)}{b - a} ] [ 1 + \sin(c) = \frac{\frac{\pi}{2} - 1 - \left( -\frac{\pi}{2} + 1 \right)}{\frac{\pi}{2} - \left( -\frac{\pi}{2} \right)} ] [ 1 + \sin(c) = \frac{\pi}{2} ]

Now, solve for ( c ) within the interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ).

[ \sin(c) = \frac{\pi}{2} - 1 ]

Given that ( -1 \leq \sin(c) \leq 1 ), ( \frac{\pi}{2} - 1 ) is within this range. Therefore, there exists a solution for ( c ) within the interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ).

To find the specific value of ( c ), take the arcsin of ( \frac{\pi}{2} - 1 ):

[ c = \arcsin\left( \frac{\pi}{2} - 1 \right) ]

This will give the value of ( c ) that satisfies the conclusion of the Mean Value Theorem on the interval ( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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