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Math Help - equation of the tangent line

  1. #1
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    Post equation of the tangent line

    Hi, I'm a little stuck on this problem. I'm not sure if I get the derivative and then use y=mx+b or what, any help much appreciated.

    Find the equation of the line tangent to the graph of the given function at the point with the indicated x coordinate.

    f(x) = x/(x^2 + 1); x=1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    Hi, I'm a little stuck on this problem. I'm not sure if I get the derivative and then use y=mx+b or what, any help much appreciated.
    i think it's nicer to use y - y_1 = m(x - x_1), but if you must use y = mx + b, that's fine.

    Find the equation of the line tangent to the graph of the given function at the point with the indicated x coordinate.

    f(x) = x/(x^2 + 1); x=1
    you start by finding the derivative of f(x), this gives a formula for the slope at any x. this will give you the value for m.

    afterwards, you need to find corresponding x and y values. you were already given the x-value. to find the y-value, plug in the given x-value into the original equation and solve for y (which is the same as f(x)). then plug in all those values into y = mx + b. b will be your only unknown, solve for it. then rewrite the equation y = mx + b, filling in only m and b, that will be the tangent line

    try it and see what you get
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    so y = 1/2?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by leviathanwave View Post
    so y = 1/2?
    yes, when x = 1, y = 1/2. so the point is (1, 1/2)

    now continue
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