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Thread: Minima problem

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    Minima problem

    THe number of minima of the polynomial 10 x^6 - 24 x^5 + 15 x^4 + 40 x^2 + 108 is ?

    (I took derivative and found a polynomial of degree 5. But making its roots and then testing each one for minima is going to be reaaaly long. Is there some easier way to do this question? )

    Thanks
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    Quote Originally Posted by champrock View Post
    THe number of minima of the polynomial 10 x^6 - 24 x^5 + 15 x^4 + 40 x^2 + 108 is ?

    (I took derivative and found a polynomial of degree 5. But making its roots and then testing each one for minima is going to be reaaaly long. Is there some easier way to do this question? )

    Thanks
    Do you know the second derivative test?

    If, at a stationary point $\displaystyle x = a, f''(a)>0$, then $\displaystyle a$ is a minimum.

    So

    $\displaystyle f'(x) = 60x^5 - 120x^4 + 60x^3 + 80x$.

    Find the stationary points...

    $\displaystyle 0 = 60x^5 - 120x^4 + 60x^3 + 80x$

    $\displaystyle 0 = 20x(3x^4 - 6x^3 + 4)$

    I'll test one of the stationary points for you... Clearly $\displaystyle x= 0$ is a stationary point.

    So let's take the second derivative, and see what it's sign is when $\displaystyle x = 0$.


    $\displaystyle f''(x) = 300x^4 - 480x^3 + 180x^2 + 80$

    $\displaystyle f''(0) = 300(0)^4 - 480(0)^3 + 180(0)^2 + 80$

    $\displaystyle f''(0) = 80 > 0$


    Therefore $\displaystyle x = 0$ is a minimum.

    Test any other stationary points in the same manner.
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