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^2-x+2# on the interval [1,3]?

Answer 1
The conclusion of the Mean Value Theorem says that there is a number #c# in the interval #(1, 3)# such that: #f'(c)=(f(3)-f(1))/(3-1)#
To find (or try to find) #c#, set up this equation and solve for #c#. If there's more than one #c# make sure you get the one (or more) in the interval #(1, 3)#.
For #f(x)--2x^2-x+2#, we have #f(1)=-1#, and #f(3)=-18-3+2=-19# Also, #f'(x)=-4x-1#.
So the #c# we're looking for satisfies:
#f'(c)=-4c-1=(f(3)-f(1))/(3-1)=(-19--1)/(3-1)=(-18)/2=-9#

So we need

#-4c-1=-9#. And #c=2#.
Note: I hope you've been told that actually finding the value of #c# is not a part of the Mean Value Theorem. The additional question"find the value of #c#" is intended as a review of your ability to solve equations. For most functions, you will not be able to find the #c# that the MVT guarantees us is there..
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Answer 2

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:

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

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

  3. 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 ]

  4. 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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Answer from HIX Tutor

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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