Вопрос школьника по предмету Алгебра
если производная функции f x равна (х2 -1)(х2-9)(х2-16) то найдите сумму длин промежутков убывания функции
Ответ учителя по предмету Алгебра
Найдем нули производной функции:
(х²
-1)(х²
-9)(х²
-16) =0
разложим каждую скобку по формуле разности квадратов: а²-в²=(а-в)(а+в)
(x-1)(x+1)(x-3)(x+3)(x-4)(x+4)=0
произведение равно нулю, когда один из множителей равен нулю
то есть корни: -1; 1; -3; 3; -4; 4
воспользуемся методом интервалов, чтобы определить знаки промежутков:
+++(-4)—(-3)+++(-1)—(1)+++(3)—(4)+++>
там где производная отрицательна, сама функция убывает
то есть нам нужны промежутки:
(-4;-3); (-1;1); (3;4)
чтобы найти длину промежутка, нужно из конечной точки вычесть начальную:
-3-(-4)=-3+4=1
1-(-1)=1+1=2
4-3=1
сумма длин промежутков: 1+2+1=4
ответ: 4
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