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Math Help - a question about tangent lines and slopes

  1. #1
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    a question about tangent lines and slopes

    the question is:
    Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.

    i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:

    (3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
    i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by mHadad View Post
    the question is:
    Find the points on the graph of y = x ^ (3/2) - x ^ (1/2) at which the tangent line is parallel to the line y - x = 3.

    i have derived the first function to find the tangent line then i have derived it again to find it's slope, then derived the second function to get it's slope, then i have set the first slope = the second slope anyway this is what i got lastly:

    (3/4) x ^ (-1/2) + (1/4) x ^ (-3/2) = 1
    i could not get the value of x from this equation.is there any formula that will get me the value of x like the quadratic formula?
    The tangent line to y=x^(3/2)-x^(1/2) is parallel to y-x=3, when the
    gradient is equal to that of the line.

    The gradient of the line is 1, so you want the solutions of:

    dy/dx=1.

    RonL
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  3. #3
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    thanks Captain for ur reply i have solved the question again and got the correct answear.anyway what if the derivative of the first function contained variable powers like this for example:
    9/5 x^(4/5) - 8/3 x^(5/3) = 1
    how can i find the value(s) of x i mean as there are two terms at the left side that contains different powers.(are there any formulas like the quadratic formula that will get me the value of x but with variable powers other than 2)??
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