# How do you determine all values of c that satisfy the mean value theorem on the interval [2,5] for #f(x) = 1 / (x-1)#?

Explanation below.

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To determine all values of ( c ) that satisfy the Mean Value Theorem on the interval ([2,5]) for ( f(x) = \frac{1}{x-1} ), we first find the average rate of change of ( f ) on the interval ([2,5]). This is calculated as:

[ f'(c) = \frac{f(5) - f(2)}{5 - 2} ]

[ f'(c) = \frac{\frac{1}{5-1} - \frac{1}{2-1}}{5-2} ]

[ f'(c) = \frac{\frac{1}{4} - 1}{3} ]

[ f'(c) = \frac{-\frac{3}{4}}{3} ]

[ f'(c) = -\frac{1}{4} ]

Now, we find the derivative of ( f(x) ):

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

To find ( c ), we set ( -\frac{1}{{(c-1)}^2} = -\frac{1}{4} ):

[ -\frac{1}{{(c-1)}^2} = -\frac{1}{4} ]

[ {(c-1)}^2 = 4 ]

[ c-1 = \pm 2 ]

[ c = 1 \pm 2 ]

Therefore, ( c ) can be ( 3 ) or ( -1 ). These are the values of ( c ) that satisfy the Mean Value Theorem on the interval ([2,5]) for ( f(x) = \frac{1}{x-1} ).

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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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