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Math Help - Calculus "Proof"

  1. #1
    Senior Member polymerase's Avatar
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    Calculus "Proof"

    Prove that there exist three different tangent lines to the curve y=x^3 throught the point (2,3)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by polymerase View Post
    Prove that there exist three different tangent lines to the curve y=x^3 throught the point (2,3)
    recall that the derivative gives the slope of the tangent line to a curve at some point.

    now, the derivative here is y' = 3x^2 we need to show that there are three such lines with this slope touching our curve.

    now, any point on our curve is given by: \left( x, x^3 \right)

    thus, using \frac {y_2 - y_1}{x_2 - x_1}, the slope of any line touching our curve and passing through the point (2,3) is given by:

    \frac {3 - x^3}{2 - x}

    now, if this slope is equal to the slope of our tangent line, we will have a tangent for every x-value given by that formula. thus we want to solve for all x's such that:

    \frac {3 - x^3}{2 - x} = 3x^2

    i leave the rest to you. showing that we have 3 solutions to this equation shows that there will be 3 tangent lines to our curve that pass through (2,3)
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by Jhevon View Post
    recall that the derivative gives the slope of the tangent line to a curve at some point.

    now, the derivative here is y' = 3x^2 we need to show that there are three such lines with this slope touching our curve.

    now, any point on our curve is given by: \left( x, x^3 \right)

    thus, using \frac {y_2 - y_1}{x_2 - x_1}, the slope of any line touching our curve and passing through the point (2,3) is given by:

    \frac {3 - x^3}{2 - x}

    now, if this slope is equal to the slope of our tangent line, we will have a tangent for every x-value given by that formula. thus we want to solve for all x's such that:

    \frac {3 - x^3}{2 - x} = 3x^2

    i leave the rest to you. showing that we have 3 solutions to this equation shows that there will be 3 tangent lines to our curve that pass through (2,3)
    I get it thanks! Just one question. I'm getting 2x^3-6x^2+3=0 and with a calculator I can easily see there are 3 (maybe irrational numbers). But during my test I'm not allowed a calculator so then how do I show there are three answers since I can't factor?
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    Quote Originally Posted by polymerase View Post
    2x^3-6x^2+3=0 But during my test I'm not allowed a calculator so then how do I show there are three answers since I can't factor?
    Can you use Descartes’ rule of sign?
    There will be two positive roots and one negative.
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by Plato View Post
    Can you use Descartes’ rule of sign?
    There will be two positive roots and one negative.
    but doesn't that rule only tell you POTENTIALLY 2 positive roots and one negative?
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    There will be two positive roots and one negative.
    Looking at  \begin{array}{l}  f(x) = 2x^3  - 6x^2  + 3 \\ <br />
 f( - 1) < 0,\quad f(0) > 0,\quad f(1) < 0\,\& \,f(3) > 0 \\  \end{array}
    it is easy to see that the negative root is between –1 & 0, two positive roots, one between 0 & 1 and one between 1 & 3.
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