Найдите наименьшее значения производной у=9соsх+10х-8 на отрезке 0 и 3п/2

Y'(f)=-9sinx+10 -9sinx+10=0 -9sinx=-10 sinx=10/9 sinx=1,1-корней нет, так каk sinx< либо=1; y(o)=9cos0+10*0+8=9+8=17 y(3п/2)=9cos(3п/2)+10*3*(3.14/2)+8=0+47,1+8=55,1 ответ : 17

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метод интервалов.

1) х² + 5х + 4 = 0

корни -1 и -4

2) х² -6х -7 = 0

корни -1 и 7

3) х² + 8х +16 = 0

    (х + 4)² = 0

корень - 4

-∞                 (-4)           [-1]             [7]               + ∞

          +                   -               +               +           это знаки х² + 5х + 4

          +                 +               -                 +           это знаки   х² -6х -7

          +                 +               +                 +           это знаки   х² + 8х +16

iiiiiiiiiiiiiiiiiiiiiii                                   iiiiiiiiiiiiiiiiiii   это решение неравенства

ответ: х ∈(-∞ ; -4)∪[7; + ∞)