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8 August, 20:28

Let f be the function given by f (x) = x+4 (x-1) (x+3) on the closed interval [-5,5]. On which closed interval is the function f guaranteed by the Extreme Value Theorem to have an absolute maximum and an absolute minimum?

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  1. 8 August, 20:40
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    Step-by-step explanation:

    f (x) = x+4 (x^2+2x-3) = 4x^2+9x-12

    f' (x) = 8x+9

    f' (x) = 0, gives x=-9/8

    f (-5) = -5+4 (-5-1) (-5+3) = -5+4*-6*-2=43

    f (-9/8) = -9/8+4 (-9/8-1) (-9/8+3)

    =-9/8+4*-17/8*15/8

    =-9/8-255/16

    =-273/16=-17 1/16

    f (5) = 4*5^2+9*5-12=100+45-12=133

    absolute maximum=133

    absolute minimum=-17 1/16
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