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Math Help - Find points on the graph...

  1. #1
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    Find points on the graph...

    Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1
    im not sure how to solve this problem or where to start. can some one help?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by viet View Post
    Q: Find all points on the graph of y = (1/3)X^3 + X^2 - X where the tangent line has slope 1
    im not sure how to solve this problem or where to start. can some one help?
    Points on the graph of:

    y = (1/3)x^3 + x^2 - x

    where the tangen has slope 1, are the points where dy/dx=1.

    So you need to differentiate (1/3)x^3 + x^2 - x with respect to x, and then
    set the derivative equal to zero and solve the resulting equation. This will
    give the x values where the slope is 1, the y values are obtained by
    substituting these x's back into y = (1/3)x^3 + x^2 - x.

    RonL
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  3. #3
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    ok so the derivative is x^2+2x-1, x = -2, x = +sqrt(8), x = -sqrt(8)
    then plug x into y = (1/3)X^3 + X^2 - X ?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by viet View Post
    ok so the derivative is x^2+2x-1, x = -2, x = +sqrt(8), x = -sqrt(8)
    then plug x into y = (1/3)X^3 + X^2 - X ?
    The derivative is dy/dx=x^2+2x-1, this equals one so:

    x^2+2x-1=1,

    so we need the roots of:

    x^2+2x-2=0

    which are x= -1+/- sqrt(3), these are plugged into y=y = (1/3)x^3 + x^2 - x
    to find the corresponding values of y.

    RonL
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  5. #5
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    ahh ok, thank you so much CaptainBlack.

    -tom
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