Calculus

posted by .

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 the difference between the Mean Value Theorem and Rollo’s Theorem.

  • Calculus -

    Rolle's theorem is just the mean value theorem, where f(x) = 0 at both endpoints.

    What did you get to parts a and b?

  • Calculus -

    I get stuck at this and don't know how to from here.... can you help...

    Q1a) h(x)=√(x+1 ) [3,8]

    MVT=[h(b)-h(a)]/(b-a)=h'(c)

    To find h(b) and h(a), we just plug endpoints into original function
    h(b)=h(8)=√(x+1 )
    h(b)=h(8)=√(8+1 ) = 3

    h(a)=h(3)= √(3+1 ) = 2

    MVT=[3-2]/[8-3] =f^' (c)
    MVT=1/2=f^' (c)

    Next, we find the derivative for h(x)

    h'(x)=√(x+1 )

    h'(x)=(d/dx(x+1))/(2√(x+1))

    h'(x)=(1+0)/(2√(x+1))

    h'(x)=1/(2√(x+1))


    h'(c)=h'(x)
    1/2=(1+0)/(2√(x+1))

  • Calculus -

    if h(x) = √(x+1)
    h'(x) = 1/(2√(x+1))

    so, we want c where
    h'(c) = (3-2)/(8-3) = 1/5

    1/2√(x+1) = 1/5
    5 = 2√(x+1)
    √(x+1) = 5/2
    x+1 = 25/4
    x = 21/4
    and 3 < 21/4 < 8
    -----------------------------
    if k(x) = (x-1)/(x+1)
    k'(x) = 2/(x+1)^2
    k(0) = 0
    k(4) = 3/5

    so, we want c where k'(c) = (3/5)/4 = 3/20

    3/20 = 2/(x+1)^2
    3(x+1)^2 = 40
    (x+1)^2 = 40/3
    x = -1 + 2√(10/3) = 2.65
    0 < 2.65 < 4, so we're ok.

  • Calculus -

    actually, I think Rolle's Theorem is the MVT where f(a) = f(b), so that f'(c) = 0.

  • Calculus -

    oops. k(0) = -1

    adjust the calculation accordingly.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Caluclus

    [Mean Value Theorem] f(x)=-3x^3 - 4x^2 - 2x -3 on the closed interval [0,8]. Find the smallest value of c that satisfies the conclusion of the Mean Value Theorem for this function defined on the given interval. I got 8 - sqrt(5696) …
  2. calculus

    Verify that the hypotheses of the Mean-Value Theorem are satisfied on the given interval, and find all values of c in that interval that satisfy the conclusion of the theorem. f(x)=x^2-3x; [-2,6]
  3. calculus

    Verify that the Intermediate Value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. f(x) = x^2 - 6x + 8, [0,3], f(c) = 0 I have no idea how to use the theorem :(
  4. calculus

    verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4]
  5. math

    verify that the function satisfies the hypothesis of the mean value theorem on the given interval. then find all numbers c that satisfy the conclusion of the mean value theorem. f(x) = x/(x+2) , [1,4]
  6. Calculus

    Verify that the hypotheses of the Mean-Value Theorem are satisfied for f(x) = √(16-x^2 ) on the interval [-4,1] and find all values of C in this interval that satisfy the conclusion of the theorem.
  7. math

    Verify that f(x) = x^3 − 2x + 6 satisfies the hypothesis of the Mean-Value Theorem over the interval [-2, 3] and find all values of C that satisfy the conclusion of the theorem.
  8. math - very urgent !

    Verify that f(x) = x^3 − 2x + 6 satisfies the hypothesis of the Mean-Value Theorem over the interval [-2, 3] and find all values of C that satisfy the conclusion of the theorem.
  9. Calculus

    The function defined below satisfies the Mean Value Theorem on the given interval. Find the value of c in the interval (1, 2) where f'(c)=(f(b) - f(a))/(b - a). f(x) = 1.5x-1 + 1.1 , [1, 2] Round your answer to two decimal places.
  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