Results 1 to 7 of 7

Math Help - Derivative, where am I going wrong in this problem?

  1. #1
    Member Marconis's Avatar
    Joined
    Sep 2010
    From
    NYC
    Posts
    118

    Derivative, where am I going wrong in this problem?

    We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

    So, it gives y= (x-1)/(x-2) on coordinate (3,2)

    I say:
    a=3
    f(a)=2
    f`(3)= ?

    Using lim h-->0 f(a+h)-f(a) / h to find f`(3)

    I plug in [(3+h-1)/(3+h-2) -2] / h

    =(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
    =3+h-1-6-2h+4 / 3h+h^2-2h
    =-2 / 2h+h-2
    =-2/-2
    =1

    The book is showing the derivative as -1. Where did I mess up here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by Marconis View Post

    I plug in [(3+h-1)/(3+h-2) -2] / h
    This is not complete

    f'(x)= \frac{f(x+h)-f(x)}{h}= \frac{\frac{x-1+h}{x-2+h}-\frac{x-1}{x-2}}{h}

    Then sub x=3
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    \displaystyle f'(3) = \lim_{x \to 3} \frac{f(x)-f(3)}{x-3}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{\frac{x-1}{x-2} - 2}{x-3}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{(x-1) - 2(x-2)}{(x-3)(x-2)}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{3-x}{(x-3)(x-2)}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{-1}{x-2} = -1
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Marconis's Avatar
    Joined
    Sep 2010
    From
    NYC
    Posts
    118
    Quote Originally Posted by skeeter View Post
    \displaystyle f'(3) = \lim_{x \to 3} \frac{f(x)-f(3)}{x-3}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{\frac{x-1}{x-2} - 2}{x-3}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{(x-1) - 2(x-2)}{(x-3)(x-2)}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{3-x}{(x-3)(x-2)}

    \displaystyle f'(3) = \lim_{x \to 3} \frac{-1}{x-2} = -1
    That is how the book shows it, but my professor taught us using the formula that I stated in my original post. Despite this, she informed us that they are the same exact things. So, doing it my way, where am I going wrong?
    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
    Quote Originally Posted by Marconis View Post
    We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

    So, it gives y= (x-1)/(x-2) on coordinate (3,2)

    I say:
    a=3
    f(a)=2
    f`(3)= ?

    Using lim h-->0 f(a+h)-f(a) / h to find f`(3)

    I plug in [(3+h-1)/(3+h-2) -2] / h

    =(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
    =3+h-1-6-2h+4 / 3h+h^2-2h
    = -h/[h(h+1)]
    = -1/(h+1)
    = -1


    your arithmetic is sloppy
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Marconis's Avatar
    Joined
    Sep 2010
    From
    NYC
    Posts
    118
    EDIT

    Got it! Thanks. I wrote it more neatly and I was able to get it, was also stupidly distributing incorrectly. Thanks again.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,541
    Thanks
    1394
    Quote Originally Posted by Marconis View Post
    We need to find the equation of a tangent line, and to do that you need to find derivative to be able to complete the equation.

    So, it gives y= (x-1)/(x-2) on coordinate (3,2)

    I say:
    a=3
    f(a)=2
    f`(3)= ?

    Using lim h-->0 f(a+h)-f(a) / h to find f`(3)

    I plug in [(3+h-1)/(3+h-2) -2] / h

    =(3+h-1)-2(3+h-2) / h(3+h-2) (After multiplying by CD)
    =3+h-1-6-2h+4 / 3h+h^2-2h
    In the numerator, 3- 1- 6+ 4= 0 and h- 2h= -h. In the denominator 3h+ h^2- 2h= h+ h^2= h(1+ h)

    =-2 / 2h+h-2
    so I don't see how you get this. It should be -1/(1+ h).

    =-2/-2
    =1

    The book is showing the derivative as -1. Where did I mess up here?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] why is this derivative wrong?
    Posted in the Calculus Forum
    Replies: 13
    Last Post: October 3rd 2010, 01:23 PM
  2. Replies: 4
    Last Post: January 28th 2010, 09:32 PM
  3. Replies: 1
    Last Post: November 8th 2009, 08:17 AM
  4. Replies: 16
    Last Post: August 5th 2009, 10:59 AM
  5. Replies: 4
    Last Post: November 8th 2008, 05:30 AM

Search Tags


/mathhelpforum @mathhelpforum