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    Part b ii) says show that the x-co-ordinates of any stationary points on the curve (y = x^4 -8x^2 + 60x + 7) satisfy the equation x^3 - 4x + 15 =0.

    That bit seemed OK

    Then part iii) said Use the results above to show the only stationary point on the curve occurs when x = -3.

    The answers that I have seen on the forum for the exam starts to use b^2-4ac and to show there is no real solution.

    But isn't that for finding roots? I've never used it in the context of stationary points. Has anyone got a resource that I could look at/visit to explain that bit to me please?
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    You've already shown that (x, f(x)) is a stationary point if and only if x^3 - 4x + 15 = 0.
    What does this tell you about when x=-3?
    And what information does that give you about what other roots of x^3 - 4x + 15 exist?
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    Oh I get it.

    Roots are solutions.

    I might not have explained properly what I'm actually thinking but it now makes sense in my head.

    Thanks
 
 
 
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