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  1. use the product rule: second der: -3e^(-3x)*sinx + e^(-3x)*cosx now inflection pt means that at x=c, second der. is 0 so: -3e^(-3c)*sin(c) + e^(-3c)*cos(c)=0 now solve this equation e^(-3c)*cos(c) = 3e^(-3c)*sin(c) we can cancel -3e^(-3c) since it occurs on both sides cos(c) = 3 sin(c...

    2 Answers · Education & Reference · 01/04/2009

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