How do you give a value of c that satisfies the conclusion of the Mean Value Theorem for Derivatives for the function #f(x)=2x^2x+2# on the interval [1,3]?
So we need
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To find the value of ( c ) that satisfies the conclusion of the Mean Value Theorem for Derivatives for the function ( f(x) = 2x^2  x + 2 ) on the interval ([1,3]), follow these steps:

Find the derivative of the function ( f(x) ). [ f'(x) = 4x  1 ]

Evaluate ( f'(x) ) at the endpoints of the interval ([1,3]). [ f'(1) = 4(1)  1 = 5 ] [ f'(3) = 4(3)  1 = 13 ]

Apply the Mean Value Theorem, which states that there exists a ( c ) in the open interval ((1,3)) such that ( f'(c) ) is equal to the average rate of change of ( f ) over the interval ([1,3]). [ f'(c) = \frac{f(3)  f(1)}{3  1} = \frac{18  (1)}{3  1} = \frac{17}{2} = 8.5 ]

Set ( f'(c) ) equal to the average rate of change found in step 3 and solve for ( c ). [ 4c  1 = 8.5 ] [ 4c = 7.5 ] [ c = \frac{7.5}{4} = 1.875 ]
Therefore, ( c = 1.875 ) satisfies the conclusion of the Mean Value Theorem for Derivatives for the given function on the interval ([1,3]).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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