# How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x)=x^2 - 2x + 5# on the interval #[1, 3]#?

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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 - 2x + 5 ) on the interval ([1, 3]), follow these steps:

- Find the derivative of ( f(x) ), denoted as ( f'(x) ).
- Evaluate ( f'(x) ) at each endpoint of the interval ([1, 3]) to find the values of ( f'(1) ) and ( f'(3) ).
- Compute the average rate of change of ( f(x) ) over the interval ([1, 3]) using the formula: [ \text{Average rate of change} = \frac{f(3) - f(1)}{3 - 1} ]
- By the Mean Value Theorem, there exists a number ( c ) in the interval ([1, 3]) such that ( f'(c) ) equals the average rate of change computed in step 3.
- Set ( f'(c) ) equal to the average rate of change and solve for ( c ).

Let's go through the steps:

- ( f'(x) = 2x - 2 )
- Evaluate:
- ( f'(1) = 2(1) - 2 = 0 )
- ( f'(3) = 2(3) - 2 = 4 )

- Average rate of change: [ \text{Average rate of change} = \frac{f(3) - f(1)}{3 - 1} = \frac{(3^2 - 2(3) + 5) - (1^2 - 2(1) + 5)}{3 - 1} = \frac{9 - 6 + 5 - 1 + 2 + 5}{2} = \frac{8}{2} = 4 ]
- Since ( f'(1) = 0 ) and ( f'(3) = 4 ), the function ( f(x) ) is continuous and differentiable on the interval ([1, 3]).
- According to the Mean Value Theorem: [ f'(c) = \frac{f(3) - f(1)}{3 - 1} = 4 ] Solve for ( c ): [ 2c - 2 = 4 ] [ 2c = 6 ] [ c = 3 ]

Thus, ( c = 3 ) satisfies the conclusion of the Mean Value Theorem for the given function on the interval ([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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