Results 1 to 5 of 5

Math Help - Confused by alternate method of finding f'(a)

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    13

    Confused by alternate method of finding f'(a)

    I'm having a hard time using:

    f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}

    to find the derivative of f(x)=x^2-8x+9

    I need to do it using f'(a)=\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}

    and the other way. I did fine finding it with the second method and came up with f'(a) = 2a-8

    Here is the order that I've tried this with the first method.

    f'(a)=\lim_{x \to a}\dfrac{x^2-8x+9-(a^2-8a+9)}{x-a}

    =\lim_{x \to a}\dfrac{x^2-8x+9-a^2-8a+9}{x-a}

    =\lim_{x \to a}\dfrac{x^2-8x-a^2+8a}{x-a}

    =\lim_{x \to a}\dfrac{x(x-8)-a(a+8)}{x-a}

    I'm not sure what to do at this point. Did I mess up and can some how use factor by grouping and then cancel out (x-a) and use substitution to get my derivative?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    You want to write the numerator as x^2-a^2-8(x-a). Then you can factor out x-a and cancel with the denominator.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,804
    Thanks
    1576
    Quote Originally Posted by Latvija13 View Post
    I'm having a hard time using:

    f'(a)=\lim_{x \to a}\dfrac{f(x)-f(a)}{x-a}

    to find the derivative of f(x)=x^2-8x+9

    I need to do it using f'(a)=\lim_{h \to 0}\dfrac{f(a+h)-f(a)}{h}

    and the other way. I did fine finding it with the second method and came up with f'(a) = 2a-8

    Here is the order that I've tried this with the first method.

    f'(a)=\lim_{x \to a}\dfrac{x^2-8x+9-(a^2-8a+9)}{x-a}

    =\lim_{x \to a}\dfrac{x^2-8x+9-a^2-8a+9}{x-a} Your mistake is here

    =\lim_{x \to a}\dfrac{x^2-8x-a^2+8a}{x-a}

    =\lim_{x \to a}\dfrac{x(x-8)-a(a+8)}{x-a}

    I'm not sure what to do at this point. Did I mess up and can some how use factor by grouping and then cancel out (x-a) and use substitution to get my derivative?
    That step should be \displaystyle \lim_{x \to a}\frac{x^2 - 8x + 9 - a^2 + 8a - 9}{x - a}, since you are subtracting EVERYTHING in that bracket...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2010
    Posts
    13
    Quote Originally Posted by DrSteve View Post
    You want to write the numerator as x^2-a^2-8(x-a). Then you can factor out x-a and cancel with the denominator.
    Don't I end up with just f'(a) = -8 doing that?

    =\lim_{x \to a}\dfrac{x^2-8x+9-a^2+8a-9}{x-a}

    =\lim_{x \to a}\dfrac{x^2-a^2-8x+8a}{x-a}

    =\lim_{x \to a}\dfrac{x^2-a^2-8(x+a)}{x-a}

    I'm not sure how I get -8(x+a) to turn into -8(x-a) But to continue.

    =\lim_{x \to a}\dfrac{x^2-a^2-8(x-a)}{x-a}

    =\lim_{x \to a}{x^2-a^2-8

    =a^2-a^2-8

    =-8

    Hate being bad at math :/

    Edit: Actually, I think it was my subtraction error that got me.

    Edit: I think I see how its done now.

    =\lim_{x \to a}\dfrac{x^2-a^2-8(x+a)}{x-a}

    I'm not sure how I get -8(x+a) to turn into -8(x-a) But to continue.
    Edit: I see now, -8(x-a) has be like that in order to get a positive a.

    =\lim_{x \to a}\dfrac{(x+a)(x-a)-8(x-a)}{x-a}

    Cancel out my two (x-a)'s

    Then I end up with:

    =\lim_{x \to a}{(x+a)-8

    ={(a+a)-8

    =2a-8
    Last edited by Latvija13; February 12th 2011 at 08:11 PM. Reason: Maybe I see the light?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Your computation after the last edit is correct.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: August 31st 2010, 01:39 PM
  2. Replies: 3
    Last Post: March 6th 2010, 04:40 AM
  3. Simplex method, confused with how to set up.
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: December 15th 2009, 07:48 AM
  4. Confused with the Lagrange method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 1st 2009, 06:46 AM
  5. Confused of Volume of Solids(Washers Method)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 22nd 2009, 07:45 PM

Search Tags


/mathhelpforum @mathhelpforum