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Math Help - deduce values of k for which cubic has three distinct real roots

  1. #1
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    deduce values of k for which cubic has three distinct real roots

    Find the equation at the tangent to  y = x^3 at the point  (t,t^3) .

    Hence deduce the values of k for which the equation  x^3 = kx -2 will have 3 real and distinct roots

    The equation of the tangent is  y = 3t^2x - 2t^3

    what do I do next?

    Thank you
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  2. #2
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    For what values of x is the tangent line to x^3- kx horizontal? If there are two such values and the value of y at one of those points is greater than 2 and at the other less than two, then the equation will have three distince real roots. Do you see why?
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  3. #3
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    sorry. I don't see why
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  4. #4
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    Well, do you know what the graph of a general cubic looks like? In general it looks like an "S" on its side- that is, the graph rises to some maximum value, then back down to a minimum value, then back up again (or the other way, first a minimum, then up to a maximu, then down again).

    In either case, the maximum value and minimum values will be where the derivative is 0. If the maximum is above y= -2 and the minimum below it, the graph must cross y= -2 going up to the maximum, then cross it again going down to the minimum, and cross it a third time going up again.

    In particular, y= x^3- kx has derivative y'= 3x^2- k which will be 0 at x= \pm\sqrt{\frac{k}{3}. At x= \sqrt{k}{3} y= \left(\frac{k}{2}\right)^{3/2}- \frac{k^{3/2}}{\sqrt{3}}= k^{3/2}\left(\frac{1}{2^{3/2}- \frac{1}{\sqrt{3}}\right). \frac{1}{2^{3/2}- \frac{1}{\sqrt{3}} is about -.224. That value of y will be greater than 2 for -.224k^{3/2}> 2 or k^{3/2}<  -8.9 or k< - 4.3
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