# Given the function #f(x)=sqrt(2-x)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-7,2] and find the c?

Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval.

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To determine whether the function ( f(x) = \sqrt{2 - x} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([-7, 2]), we need to check if the following conditions are met:

- ( f(x) ) is continuous on ([-7, 2]).
- ( f(x) ) is differentiable on ((-7, 2)).

( f(x) = \sqrt{2 - x} ) is continuous everywhere except at ( x = 2 ) because the square root function is continuous for all real numbers except when its argument is negative.

To check differentiability, we need to ensure that the derivative exists for all ( x ) in ((-7, 2)). Taking the derivative of ( f(x) ):

[ f'(x) = \frac{d}{dx} \sqrt{2 - x} = \frac{1}{2\sqrt{2 - x}} ]

( f'(x) ) exists for all ( x ) in ((-7, 2)) except at ( x = 2 ) where ( f(x) ) is not differentiable. However, this point is not in the interval ([-7, 2]), so it does not affect the applicability of the Mean Value Theorem on this interval.

Since both conditions are satisfied, we can apply the Mean Value Theorem. According to the theorem, there exists a point ( c ) in ((-7, 2)) such that:

[ f'(c) = \frac{f(2) - f(-7)}{2 - (-7)} ]

[ f'(c) = \frac{\sqrt{2 - 2} - \sqrt{2 - (-7)}}{2 - (-7)} ]

[ f'(c) = \frac{\sqrt{0} - \sqrt{9}}{9} ]

[ f'(c) = \frac{0 - 3}{9} ]

[ f'(c) = -\frac{1}{3} ]

Thus, there exists ( c ) in ((-7, 2)) such that ( f'(c) = -\frac{1}{3} ).

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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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