posted by .

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.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. math- mean value theorem

    Hi I am having some trouble with these few quetions I would appreciate some help so that I can understand them better. 1) What, if anything, does the mean value theorem guarantee for the given function on this interval?
  2. calculus

    lim t-->0 h(t), where h(t)= cos(t)-1/t^2. for the table values t is +/- .002 , .0001 , .00005 , .00001 what do you guys get for the value of h(t) and limit?
  3. calculus

    let f(x)= (x-3)^-2 Show that there is no value of c in (1,4) such that f(4)-f(1)= (f prime of c)(4-1). Why doesn't this contradict the mean value theorem.
  4. calculus

    let f(x)= 2 - |2x-1|. Show that there is no value of c such that f(3)- f(0) = f'(c)(3-0). Why does this not contradict the mean value theorem.
  5. Calculus

    Let f be a twice-differentiable function such that f(2)=5 and f(5)=2. Let g be the function given by g(x)= f(f(x)). (a) Explain why there must be a value c for 2 < c < 5 such that f'(c) = -1. (b) Show that g' (2) = g' (5). Use …
  6. Calculus

    Verify the hypothesis of the mean value theorem for each function below defined on the indicated interval. Then find the value “C” referred to by the theorem. Q1a) h(x)=√(x+1 ) [3,8] Q1b) K(x)=(x-1)/(x=1) [0,4] Q1c) Explain …
  7. Calculus

    Suppose f(x)= 7-8x^2, by the Mean Value Theorem, we know there exists a c in the open interval (-2,5) such that f'(c) is equal to this mean slope. For this problem, there is only one c that works. Find it.
  8. Calculus

    show that ((x − 1)/x) <( ln x) < (x − 1) for all x>1 Hint: try to apply the Mean Value Theorem to the functions f(x) = lnx and g(x) = xlnx. I'm having trouble applying the mean value theorem
  9. Math (Calculus) (mean value theorem emergency)

    Consider the graph of the function f(x)=x^2-x-12 a) Find the equation of the secant line joining the points (-2,-6) and (4,0). I got the equation of the secant line to be y=x-4 b) Use the Mean Value Theorem to determine a point c in …
  10. Calculus

    Let f(x)=αx^2+βx+γ be a quadratic function, so α≠0, and let I=[a,b]. a) Check f satisfies the hypothesis of the Mean Value Theorem. b)Show that the number c ∈ (a,b) in the Mean Value Theorem is the midpoint of the interval …

More Similar Questions