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calculus

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find max and min of e^x-e^(3x) on interval where x is greater than or equal to 0 and less than or equal to 1

  • calculus -

    Since e^3x > e^x when x > 0,

    min at x=0: e^0 - e^0 = 0
    max at x=1: e^3 - e^1

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