Найдите наибольшее значение функции y=(x+4)^2 (x+1)+19 на отрезке [-5;-3]

Найдите наибольшее значение функции y=(x+4)^2 (x+1)+19 на отрезке [-5;-3]
- Предмет:
  Алгебра
- Автор:
  noahk18g
- 5 лет назад

Ответы 1

y=(x+4)²*(x+1)+19 [-5;-3]y`=((x+4)²)`*(x-1)+(x+4)²*(x+1)`=02(x+4)*(x+1)+(x+4)²=02x²+10x+8+x²+8x+16=03x²+18x+24=0 |÷3x²+6x+8=0 D=4x₁=-4 x₂=-2y(-5)=(-5+4)²*(-5+1)+19=(-1)²1*(-4)+19=15y(-4)=(-4+4)²*(-4+1)+19=0+19=19=ymaxy(-3)=(-3+4)²*(-3+1)+19=1²*(-2)+19=17
- Автор:
  amare
- 5 лет назад
- 0
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Найдите наибольшее значение функции y=(x+4)^2 (x+1)+19 на отрезке [-5;-3]

Ответы 1

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