Results 1 to 7 of 7

Math Help - Help finding the tangent function??

  1. #1
    Junior Member
    Joined
    Sep 2010
    Posts
    51

    Help finding the tangent function??

    Let f(x) = (3x^2)/(2x+1). For what values of x does f(x) have a tangent with slope 4/3 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2010
    Posts
    40
    Quote Originally Posted by yess View Post
    Let f(x) = (3x^2)/(2x+1). For what values of x does f(x) have a tangent with slope 4/3 ?
    Slope of the tangent = gradient of tangent=gradient of curve = dy / dx = 4/3

    Find dy / dx, for y = f(x), and then form the equation: dy / dx=4/3.
    Solve for x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2010
    Posts
    51
    Quote Originally Posted by Ithaka View Post
    Slope of the tangent = gradient of tangent=gradient of curve = dy / dx = 4/3

    Find dy / dx, for y = f(x), and then form the equation: dy / dx=4/3.
    Solve for x.
    aaah i keep getting stuck! i get this big long equation and i dont know how to simplify it! i got:
    lim of Δx->0 [(6xΔx + 6xΔx + 6xΔx + 3Δx)/(4x + 8xΔx + 4x + 1)]/Δx
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2010
    Posts
    1
    You don't need to find the derivative by first principles. Just use the product rule.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1630
    Or even the quotient rule!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1630
    Quote Originally Posted by yess View Post
    aaah i keep getting stuck! i get this big long equation and i dont know how to simplify it! i got:
    lim of Δx->0 [(6xΔx + 6xΔx + 6xΔx + 3Δx)/(4x + 8xΔx + 4x + 1)]/Δx
    This is the same as
    \lim_{\Delta x\to 0}\frac{6x^2\Delta x+ 6x (\Delta x)^2+ 6x\Delta x+ 3(\Delta x)^3}{(4x^2+ 8x\Delta x+ 4x+ 1)\Delta x}

    Since there is a " \Delta x" in each term in the numerator, you can cancel that " \Delta x" multiplying the denominator:
    \lim_{\Delta x\to 0}\frac{6x^2+ 6x\Delta x+ 6x+ 3(\Delta x)^3}{4x^2+ 8x\Delta x+ 4x+ 1}

    Now, neither numerator nor denominator goes to 0 as \Delta x goes to 0.
    Last edited by HallsofIvy; October 5th 2010 at 04:54 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2010
    Posts
    51
    Quote Originally Posted by HallsofIvy View Post
    This is the same as
    \lim_{\Delta x\to 0}\frac{6x^2\Delta x+ 6x (/Delta x)^2+ 6x\Delta x+ 3(\Delta x)^3}{(4x^2+ 8x\Delta x+ 4x+ 1)\Delta x}

    Since there is a " \Delta x" in each term in the numerator, you can cancel that " \Delta x" multiplying the denominator:
    \lim_{\Delta x\to 0}\frac{6x^2+ 6x\Delta x+ 6x+ 3(\Delta x)^3}{4x^2+ 8x\Delta x+ 4x+ 1}

    Now, neither numerator nor denominator goes to 0 as \Delta x goes to 0.
    so then do i sub in 0 for wherever there is a Δ then equate that to 4/3? because when i factor that out i get x= -2,1 but the question asks for THE x value as in only one??
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: October 27th 2010, 12:20 PM
  2. Finding slope of a tangent to a function
    Posted in the Calculus Forum
    Replies: 5
    Last Post: March 18th 2010, 12:24 PM
  3. Finding tangent to function parallell with line
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 2nd 2009, 12:50 AM
  4. Finding tangent line of a function
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 11th 2009, 10:10 AM
  5. Replies: 2
    Last Post: May 15th 2008, 03:28 AM

Search Tags


/mathhelpforum @mathhelpforum