Using the principle of the mean-value theorem on the indicated interval, how do you find all numbers c that satisfy the conclusion of the theorem #f(x) = ln (x-1)#; [2, 6]?

Answer 1

See below.

What does the conclusion of the Mean Value Theorem say?

There is a #c# in #(2,6)# such that
#f'(c) = (f(6)-f(2))/(6-2)#
To find the #c#. Find #f'(x)#, find #(f(6)-f(2))/(6-2)#.
Set them equal to each other and solve the equation. Any (and all) solutions in #(2,6)# are values for #c#.

Solve

#1/(x-1) = ln5/4#
#x-1 = 4/ln5#
#x=(4+ln5)/ln5 ~~ 3.485#
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Answer 2

To find all numbers ( c ) that satisfy the conclusion of the Mean Value Theorem for the function ( f(x) = \ln(x-1) ) on the interval ( [2, 6] ), follow these steps:

  1. Verify that the function ( f(x) ) is continuous on the closed interval ( [2, 6] ) and differentiable on the open interval ( (2, 6) ).
  2. Calculate the derivative of ( f(x) ) with respect to ( x ). The derivative of ( \ln(x-1) ) is ( \frac{1}{x-1} ).
  3. Apply the Mean Value Theorem, which states that there exists a number ( c ) in the open interval ( (2, 6) ) such that ( f'(c) ) equals the average rate of change of ( f(x) ) over ( [2, 6] ).
  4. Find the average rate of change of ( f(x) ) over ( [2, 6] ) by calculating ( \frac{f(6) - f(2)}{6 - 2} ).
  5. Set the derivative ( f'(c) ) equal to the average rate of change found in step 4 and solve for ( c ). The equation to solve is ( \frac{1}{c-1} = \frac{\ln(5) - \ln(1)}{6 - 2} ).
  6. Solve the equation for ( c ) to find all values of ( c ) that satisfy the conclusion of the Mean Value Theorem on the interval ( [2, 6] ).
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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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