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30 November, 02:26

What is the local minimum value of the function? (Round answer to the nearest hundredth) g (x) = x^4-5x^2+4

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  1. 30 November, 02:52
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    The local minimum of function is an argument x for which the first derivative of function g (x) is equal to zero, so:

    g' (x) = 0

    g' (x) = (x^4-5x^2+4) '=4x^3-10x=0

    x (4x^2-10) = 0

    x=0 or 4x^2-10=0

    4x^2-10=0 / 4

    x^2-10/4=0

    x^2-5/2=0

    [x-sqrt (5/2) ][x+sqrt (5/2) ]=0

    Now we have to check wchich argument gives the minimum value from x=0, x=sqrt (5/2) and x=-sqrt (5/2).

    g (0) = 4

    g (sqrt (5/2)) = 25/4-5*5/2+4=4-25/4=-9/4

    g (-sqrt (5/2)) = - 9/4

    The answer is sqrt (5/2) and - sqrt (5/2).
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