How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x)=x-2sinx# on the interval [0, pi]?
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To find the number ( c ) that satisfies the conclusion of the Mean Value Theorem for the function ( f(x) = x - 2\sin(x) ) on the interval ([0, \pi]), we first need to check if the conditions of the Mean Value Theorem are met. The conditions are:
- ( f(x) ) must be continuous on the closed interval ([a, b]).
- ( f(x) ) must be differentiable on the open interval ((a, b)).
Given that ( f(x) = x - 2\sin(x) ), it's continuous and differentiable everywhere, including the interval ([0, \pi]).
Now, to find the number ( c ), we apply the conclusion of the Mean Value Theorem, which states:
[ f'(c) = \frac{f(b) - f(a)}{b - a} ]
Substituting the values into the equation:
[ f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} ]
[ f'(c) = \frac{\pi - 2\sin(\pi) - (0 - 2\sin(0))}{\pi} ]
[ f'(c) = \frac{\pi - 0 - 0}{\pi} ]
[ f'(c) = \frac{\pi}{\pi} ]
[ f'(c) = 1 ]
Now, we need to find the derivative of ( f(x) ), which is:
[ f'(x) = 1 - 2\cos(x) ]
Setting ( f'(c) = 1 ), we have:
[ 1 - 2\cos(c) = 1 ]
[ -2\cos(c) = 0 ]
[ \cos(c) = 0 ]
The solutions to ( \cos(c) = 0 ) on the interval ([0, \pi]) are ( c = \frac{\pi}{2} ) and ( c = \frac{3\pi}{2} ). However, only ( c = \frac{\pi}{2} ) falls within the interval ([0, \pi]).
Therefore, the number ( c ) that satisfies the conclusion of the Mean Value Theorem for the function ( f(x) = x - 2\sin(x) ) on the interval ([0, \pi]) is ( c = \frac{\pi}{2} ).
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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.
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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