Results 1 to 7 of 7

Math Help - Computing Derivatives with a graph?

  1. #1
    Junior Member
    Joined
    Sep 2008
    Posts
    43

    Computing Derivatives with a graph?

    Okay Im REALLY confused with derivatives and how to compute them. THe problem I have is this:
    There is a graph in which a straight line crosses -0.25 on the x-axis and 1 on the y-axis. Find f'(1/2).

    What do I do?? Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2007
    Posts
    209
    The derivative is basically the slope.

    If you are asked to find a derivative of a certain point on the graph, just draw a tangent line to the point, pick 2 points on the tangent line and use the slope formula.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2008
    Posts
    43
    How do you find the tangent line?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2007
    Posts
    209
    Since you are only given the graph and not the function, just draw the tangent line to whichever point you need to find the derivative for.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2008
    Posts
    43
    Im sorry, I appreciate your help but im still confused. sorry if im slow so do i just draw a line on the x-axis at 1/2?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by jemadd2 View Post
    How do you find the tangent line?
    For example, I was given a graph where a straight line crossed the x-axis at -0.25 and the y-axis at 1. Find f'(1/2). So to do this I would need to find the tangent line right? Im confused
    the derivative gives you the formula for the slope. once you find the derivative and plug in the required x-value and that gives you your m in the equation of a line (y = mx + b), you can then use your m and a point the line passes through to get your tangent line. however, that is not the case here. you are required to find a slope, why would you find the tangent line. that's not what they asked for, and you don't need it to find the slope. (the tangent line for a straight line, at any point on the line, is itself, by the way). all you are doing is finding the slope of a line that passes through (-0.25, 0) and (0, 1). the slope (for all points on the straight line) is just given by m = \frac {y_2 - y_1}{x_2 - x_1}, and you know what those components are, right?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,691
    Thanks
    449
    the equation of the described line is y = 4x+1 ...

    but everything else you've stated seems disconnected without additional information on what you were given and what you are trying to accomplish.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Computing ROI ?
    Posted in the Business Math Forum
    Replies: 1
    Last Post: May 13th 2009, 02:46 PM
  2. How To Graph Derivatives?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 25th 2009, 04:38 AM
  3. Computing e^A
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: November 27th 2008, 06:49 AM
  4. derivatives with a graph
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 10th 2007, 08:09 PM
  5. computing
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: April 28th 2007, 12:25 AM

Search Tags


/mathhelpforum @mathhelpforum