Results 1 to 5 of 5

Math Help - consider the function using Mean Value Theorem

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    34

    consider the function using Mean Value Theorem

    Consider the function f(x) = e^(5x/31) , where the interval is 4 <= x <= 6

    (a) We know that the Mean Value Theorem can be applied to (f) on the interval [4,6] because (f) is ______ on [4,6] and _____ on (4,6).
    Note. You are to enter two mathematical terms as answers.

    (b) If we now apply the Mean Value Theorem to (f) on the interval [4,6], we can conclude that there is at least one number (c) in the interval (4,6) such that f'(c) =_______.

    (c) Find all possible value(s) for such (c)



    Okay so i know that the MVT says that a continuous function on [a,b] an differentiable on (a,b) where a <b then there exists a c value in (a,b) such that f'(c) = [f(b) - f(a)] / (b-a)

    I entered my a and bvalues into teh derivative equation as given in the MVT,
    f'(x) = [(e^(5*4/31)) - (e^(5*6/31))] / (4-6)

    --> f'(x) = [(e^(20/31)) - (e^(30/31))] / (-2)
    but i just don't know where to go from here..
    do i equate it to zero?
    this is a new concept for us, please help

    thanks in advance,
    britt w
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by williamb View Post
    Consider the function f(x) = e^(5x/31) , where the interval is 4 <= x <= 6 ...

    ... f'(x) = [(e^(20/31)) - (e^(30/31))] / (-2)
    what is f'(x) ?

    once you find it, set it equal to the average rate of change (the right side of the above equation) , then solve for the value of x in the interval (0,4) that is guaranteed to exist by the MVT.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    34
    do you mean find the actual NUMBER value of the derivative?
    how do i calculate that?
    [(e^(20/31)) - (e^(30/31))] / (-2)

    when i press "e^x" on my calculator it just returns 1
    so do i have to enter it 20/31 e^x for the correct answer?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    34
    k this is where i'm at..

    so i take my first derivative (slope of secant line), which is on the RHS
    and i equate it to the slope of my tangent line, which is on the LHS

    (e^(5x/31)*5/31) = (e^(20/31)-e^(30/31))/-2

    i need to solve for x, i THINK i may have gotten it.. so far.. who knows:

    (e^(5x/31)*5/31) = -0.7257000/-2
    (e^(5x/31)*5/31) = 0.362850029
    e^(5x/31) = 2.249670184
    take the ln of each to bring the exponent down, and solve
    5x/31 = ln(2.249670184)
    x = [31 * ln(2.249670184)]/5

    so now i...
    plug x back into (e^(5x/31)*5/31) = (e^(20/31)-e^(30/31))/-2 to get my f'(x)??

    (e^(5([31 * ln(2.249670184)]/5)/31)*5/31) = (e^(20/31)-e^(30/31))/-2

    in the exponent i suppose we could simplify to get

    (e^(ln(2.249670184))*5/31) = (e^(20/31)-e^(30/31))/-2

    and then just transfer over to one side for our f'(x)???
    f'(x) = [(e^(20/31)-e^(30/31))/-2 ] - [(e^(ln(2.249670184))*5/31)]


    if this is the case and in some nuts way i'm correct, how do we find the c value?!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
     <br />
f(x) = e^{\frac{5x}{31}}<br />

     <br />
f'(x) = \frac{5}{31}e^{\frac{5x}{31}}<br />

     <br />
\frac{5}{31}e^{\frac{5x}{31}} = \frac{f(6) - f(4)}{6 - 4}<br />

     <br />
\frac{5}{31}e^{\frac{5x}{31}} = \frac{e^{\frac{30}{31}}-e^{\frac{20}{31}}}{2}<br />

     <br />
e^{\frac{5x}{31}} = \frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)<br />

     <br />
\frac{5x}{31} = \ln\left[\frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)\right]<br />

     <br />
x = \frac{31}{5} \ln\left[\frac{31}{10} \left(e^{\frac{30}{31}}-e^{\frac{20}{31}}\right)\right] \approx 5.027<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 29th 2011, 02:08 PM
  2. Implicit function theorem
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: November 29th 2010, 03:19 PM
  3. Replies: 0
    Last Post: November 13th 2009, 05:41 AM
  4. inverse function theorem?
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 6th 2009, 11:57 PM
  5. euler function and theorem
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 30th 2007, 11:59 AM

Search Tags


/mathhelpforum @mathhelpforum