Results 1 to 5 of 5

Math Help - help with calculus problem: Finding equation of tangent line at given point

  1. #1
    Newbie
    Joined
    Oct 2010
    Posts
    6

    help with calculus problem: Finding equation of tangent line at given point

    Hi, I'm new to this forum and I need some help with this problem. If you could help, I'd appreciate it.

    Find an equation of the tangent line at the point indicated.
    f (x) = log2 (3x + x^-3), at x = 1

    I tried using y - y1 = m(x - x1)

    for y1, i got 2, x1 is 1, and for the slope i got (3-3/x^4)/((1/x^3+3 x) log(2))

    so for the equation of the line, I got ((3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)) +2

    but unfortunately, its wrong.. lol
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by justin144 View Post
    Hi, I'm new to this forum and I need some help with this problem. If you could help, I'd appreciate it.

    Find an equation of the tangent line at the point indicated.
    f (x) = log2 (3x + x^-3), at x = 1

    I tried using y - y1 = m(x - x1)

    for y1, i got 2, x1 is 1, and for the slope i got (3-3/x^4)/((1/x^3+3 x) log(2))

    so for the equation of the line, I got ((3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)) +2

    but unfortunately, its wrong.. lol
    m = f'(1) = ....

    f(1) = 2 so use x1 = 1 and y1 = 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2010
    Posts
    6
    right. this is what I did. so I got :
    y-2 = (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)
    add two to both sides which gives:
    y= (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1) +2
    but the program is still counting my answer as incorrect.
    So did I get the slope wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by justin144 View Post
    right. this is what I did. so I got :
    y-2 = (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1)
    add two to both sides which gives:
    y= (3-3/x^4)/((1/x^3+3 x) log(2)) (x-1) +2
    but the program is still counting my answer as incorrect.
    So did I get the slope wrong?
    For the second time:

    m = f'(1) = ....

    Do you know what this means? It means the value of the derivative at x = 1.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2010
    Posts
    6
    Quote Originally Posted by mr fantastic View Post
    For the second time:

    m = f'(1) = ....

    Do you know what this means? It means the value of the derivative at x = 1.
    ohhh ok. so plugging x=1 into the derivative, i get:
    m=0, sooo:
    y-2 = 0(x-1)
    which is: y=2
    Ok, i'll try that. Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 30th 2011, 02:13 PM
  2. Equation of the tangent line at that point
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 13th 2011, 04:44 AM
  3. Equation of a tangent line at a certain point.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 15th 2009, 02:02 PM
  4. Replies: 2
    Last Post: October 12th 2009, 04:33 PM
  5. Finding a Tangent Line without Calculus
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: December 17th 2006, 02:24 PM

Search Tags


/mathhelpforum @mathhelpforum