Results 1 to 3 of 3

Math Help - Mean Value theorem question

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    79

    Mean Value theorem question

    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    f(x)= \frac{x}{x+2}, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    \frac{2/3-1/3}{4-1}

    \frac{1/3}{3}
    =1/9

    <br />
f'=\frac{(x+2) -x}{(x+2)^2}<br />

    <br />
f'=\frac{(2)}{(x+2)^2}<br />
 =1/9

    = (x+2)^2=18
    = (x^2+4x-14)

    Did I make a mistake or am I suppose to use the quadratic formula to get c?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member utopiaNow's Avatar
    Joined
    Mar 2009
    Posts
    72
    Thanks
    1
    Quote Originally Posted by hazecraze View Post
    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    f(x)= \frac{x}{x+2}, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    \frac{2/3-1/3}{4-1}

    \frac{1/3}{3}
    =1/9

    <br />
f'=\frac{(x+2) -x}{(x+2)^2}<br />

    <br />
f'=\frac{(2)}{(x+2)^2}<br />
 =1/9

    = (x+2)^2=18
    = (x^2+4x-14)

    Did I make a mistake or am I suppose to use the quadratic formula to get c?
    Looks fine, yes use the quadratic formula to find the c in the interval a < c < b ie. accept the c in this interval and reject the value for c which is not in this interval.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by hazecraze View Post
    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    f(x)= \frac{x}{x+2}, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    \frac{2/3-1/3}{4-1}

    \frac{1/3}{3}
    =1/9

    <br />
f'=\frac{(x+2) -x}{(x+2)^2}<br />

    <br />
f'=\frac{(2)}{(x+2)^2}<br />
=1/9

    = (x+2)^2=18
    = (x^2+4x-14)

    Did I make a mistake or am I suppose to use the quadratic formula to get c?

    So far so good, now continue: (x+2)^2=18 \Longrightarrow x+2=\pm \sqrt{18}=\pm 3\sqrt{2} \Longrightarrow  x=\left\{\begin{array}{cc}3\sqrt{2}-2\\-3\sqrt{2}-2\end{array}\right.

    Of course, you need (and can) take only one of the above values.

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Theorem Question
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 4th 2011, 11:40 PM
  2. Another mean value theorem question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 6th 2010, 03:46 AM
  3. Mean Value Theorem Question
    Posted in the Calculus Forum
    Replies: 11
    Last Post: April 27th 2010, 07:28 AM
  4. Question on Rolle's Theorem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 20th 2009, 09:26 AM
  5. Mean Value Theorem Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 11th 2007, 09:49 AM

Search Tags


/mathhelpforum @mathhelpforum