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Thread: Differentiation question

  1. #1
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    Differentiation question

    Question: Show that, when k is constant, the curve


    $\displaystyle y = 3x^4- 8x^3-6x^2+24x+k $
    has a stationary point when x =1 and find the values of x at the other two stationary points on the curve. Find the values of k for which the curve touches the x axis.


    My attempt:
    $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24 $
    when $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=12x^{3}-24x^{2}-12x+24 = 0 $


    $\displaystyle 12x^{3}-24x^{2}-12x+24 = 0 $


    $\displaystyle 12(x^{3}-2x^{2}-x+2) = 0 $
    try x = 1




    $\displaystyle 12(1^{3}-2(1)^{2}-1(1)+2) = 0 $ Thus since f'(1) =0 thus at x =1 there is a stationary point.


    The other values of x are x =-1 and x =2


    my problem is how to find the values of k


    please can you help me


    thank you
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  2. #2
    MHF Contributor
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    Re: Differentiation question

    Firstly, think about what the curve looks like.

    1. It is a quartic with three stationary points at x = -1, 1 and 2. What sort of stat pts are they, ie max, min, phi?

    2. What happens to the curve as x->infinity and x-> - infinity ie on your graph, does it "start" in the top left corner or bottom left corner? Does it end up in the top right corner or the bottom right corner?

    That should give you the general shape of the graph.

    3. What does "touches" the x-axis mean? At what points could your graph touch the x-axis?

    Can you see what to consider next?
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: Differentiation question

    Nevermind...wasn't paying attention.
    Last edited by MarkFL; Mar 1st 2019 at 07:50 PM.
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