Results 1 to 10 of 10

Math Help - Graphing a rational function

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    42

    Graphing a rational function

    Got another problem...

    Given f(x) = (x+5)/(2x-4)

    Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

    I've calulated the first and second derivatives and come up with this:

    f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

    f ''(x) = 8(x-17) / (2x-4)^3

    I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013
    Quote Originally Posted by Jools View Post
    Got another problem...

    Given f(x) = (x+5)/(2x-4)

    Calculate the first and second derivatives, the intervals on which the function is increasing and decreasing, inflection points, and concavity.

    I've calulated the first and second derivatives and come up with this:

    f '(x) = (2x^2 + 4x - 30) / (2x-4)^2

    f ''(x) = 8(x-17) / (2x-4)^3

    I'm pretty sure those are right... but maybe not. In orer to find the intervals of increase and decrease I should find where f '(x) = 0 right? If that's the case I may need a little push to get that figured out. Based on the second derivative, the inflection point should be x=17 right? That doesn't seem right either... I'm all screwed up on this one I think...
    f'(x) is incorrect.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    Quote Originally Posted by skeeter View Post
    f'(x) is incorrect.
    Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

    -14 / (2x-4)^2

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013
    Quote Originally Posted by Jools View Post
    Gotcha... I just checed f '(x) again and see where I made the errror. Does this look better for f '(x)?

    28 / (2x-4)^3

    Thanks!
    f(x) = \frac{x+5}{2x-4}

    f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}

    f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}

    f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}

    f'(x) = -\frac{7}{2(x-2)^2}


    f'(x) = -\frac{7}{2}(x-2)^{-2}

    f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    Quote Originally Posted by skeeter View Post
    f(x) = \frac{x+5}{2x-4}

    f(x) = \frac{1}{2} \cdot \frac{x+5}{x-2}

    f'(x) = \frac{1}{2} \cdot \frac{(x-2)(1) - (x+5)(1)}{(x-2)^2}

    f'(x) = \frac{1}{2} \cdot \frac{x-2 - x - 5}{(x-2)^2}

    f'(x) = -\frac{7}{2(x-2)^2}


    f'(x) = -\frac{7}{2}(x-2)^{-2}

    f''(x) = 7(x-2)^{-3} = \frac{7}{(x-2)^3}

    Here's how I came up with f '(x)... Can you show me where I went wrong?

    f(x) = (x+5) / (2x-4)

    f(x) = (x+5)(2x-4)^-1
    f '(x) = (x+5)' (2x-4)^-1 + (x+5)(2x-4)^-1'
    = (1)(2x-4)^-1 + (x+5)(-1)(2x-4)^-2(2)
    = (2x-4)^-1 - 2(x+5)(2x-4)^-2
    = 1 / (2x-4) - 2(x+5) / (2x-4)^2
    = (2x-4)/(2x-4)^2 - 2(x+5)/2x-4)^2
    = 2x-4 - 2(x+5) / (2x-4)^2
    = 2x-4-2x-10 / (2x-4)^2
    f '(x) = -14 / (2x-4)^2

    Did I go wrong somewhere? It's close to what you have... Only I've gotten -14 as the numerator instead of -7.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013
    it's the same thing, I just simplified a bit more ...

    \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    Quote Originally Posted by skeeter View Post
    it's the same thing, I just simplified a bit more ...

    \frac{-14}{(2x-4)^2} = \frac{-14}{[2(x-2)]^2} = \frac{-14}{4(x-2)^2}= \frac{-7}{2(x-2)^2}
    I don't have the 4 in the denominator though...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013
    Quote Originally Posted by Jools View Post
    I don't have the 4 in the denominator though...
    and you shouldn't ... look at the last equation again.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Feb 2010
    Posts
    42
    Quote Originally Posted by skeeter View Post
    and you shouldn't ... look at the last equation again.
    Sorry I'm missing something here...

    I'm getting

    f '(x) = -14 / (2x-4)^2

    You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,134
    Thanks
    1013
    Quote Originally Posted by Jools View Post
    Sorry I'm missing something here...

    I'm getting

    f '(x) = -14 / (2x-4)^2

    You have -14 / [2(x-2)]^2 which is working out to -14 / 4(x-2)^2. Should you not multiply out the 2(x-2) portion first before squaring it?
    working with the denominator ...

     <br />
(2x-4)^2 = [2(x-2)]^2 = 2^2(x-2)^2 = 4(x-2)^2<br />

    now put that under -14 ...

     <br />
\frac{-14}{4(x-2)^2} = \frac{-7}{2(x-2)^2} = \frac{-14}{(2x-4)^2}<br />
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Graphing a Rational Function
    Posted in the Algebra Forum
    Replies: 9
    Last Post: January 27th 2011, 12:10 PM
  2. Graphing of rational fn
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 2nd 2010, 03:25 AM
  3. Rational function graphing
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: September 28th 2009, 09:19 AM
  4. Graphing a rational function with an absolute value
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 9th 2008, 05:43 PM
  5. graphing rational function
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 28th 2006, 01:17 PM

Search Tags


/mathhelpforum @mathhelpforum