Let f(x) = (x+1)/(x-1). Show that there are no vlue of c such that

f(2)-f(0) =f'(c)(2-0). Why does this not contradict the Mean Value Theorem?

I plugged 2 and 0 into the original problem and got 3 and -1 . Then I found
the derivative to be ((x-1)-(x+1))/(x-1)^2. Whould would I do next? I am
confused at this part.

The mean value theorem says, in this example, that there IS some value of c in the interval such that f'(c) = f(2)-f(0)/2-0) = 4/2 = 2

The derivative is
f'(x) = -2/(x-1)^2

For that to equal 2, you must have
2 = -2/(x-1)^2
(x-1)^2 = -1
That is not possible since (x-1)^2 must always be positive.

The reason the Mean Value Theorm seems to be violated here is that the function f(x) is not continuous in the inveral x = 0 to 2. The Mean Value Theorem does not apply in such situations.

To show that there are no values of c such that f(2) - f(0) = f'(c)(2-0), you have correctly found the derivative of f(x), which is f'(x) = -2/(x-1)^2.

To proceed, you can substitute the values of f(2) = 3 and f(0) = -1 into the equation f(2) - f(0) = f'(c)(2-0):

3 - (-1) = f'(c)(2-0)
4 = f'(c) * 2

Now, you can solve for f'(c):

f'(c) = 4/2
f'(c) = 2

Next, you can equate f'(c) to the derivative you found:

2 = -2/(c-1)^2

To solve this equation, you can multiply both sides by (c-1)^2:

2(c-1)^2 = -2

Divide both sides by 2:

(c-1)^2 = -1

This equation leads to a contradiction because (c-1)^2 must always be positive, while -1 is not positive. Therefore, there is no value of c that satisfies this equation, which means there are no values of c such that f(2) - f(0) = f'(c)(2-0).

This contradiction does not violate the Mean Value Theorem because the function f(x) = (x+1)/(x-1) is not continuous on the interval [0, 2]. The Mean Value Theorem applies only to continuous functions, and in this case, since f(x) has a discontinuity at x = 1, the theorem does not apply.