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Igor171717
@Igor171717
September 2021
2
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№12 под цифрами 6,7,8
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Дмитрий1425
Task/27145250
-------------------
6.
y =1 -(x+1)
⁶
y ' = - 6(x+1)
⁵
y ' = 0 ⇒ x = - 1
y ' "+ " " - "
------------------ (-1) ---------------
y ↑
max ↓
----
7.
y =(x+2)²(x-3)³
* * * (uv) ' =u ' *v +u*v ' * * *
y ' =2(x+2)*(x-3)³ +(x+2)²*3(x-3)² =(x+2)(x-3)²(2x -6 +3x+6) =5x(x+2)(x-3)² ;
y ' = 0 ⇔ (x+2)x(x-3)² = 0 ⇒ x₁ = - 2 , x₂ =0 , x₃ =3 .
y ' " -" " + " " - " " + "
------------------ (-2) ----------------( 0) ---------------(3) --------------
y ↓
min
↑
max
↓
min
↑
----
8.
y =(x-5)e
ˣ
y ' = (x-5) '*eˣ +(x-5)*(eˣ) ' =1*eˣ +(x-5)*eˣ =(x-4)*eˣ
y ' = 0 ⇔(x-4)*eˣ = 0 ⇒ x =4 .
y ' " - " " + "
------------------ (4) ---------------
↓
min
↑
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oganesbagoyan
Verified answer
',".',".',".',".".".'."".."".
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Answers & Comments
-------------------
6.
y =1 -(x+1)⁶
y ' = - 6(x+1)⁵
y ' = 0 ⇒ x = - 1
y ' "+ " " - "
------------------ (-1) ---------------
y ↑ max ↓
----
7.
y =(x+2)²(x-3)³
* * * (uv) ' =u ' *v +u*v ' * * *
y ' =2(x+2)*(x-3)³ +(x+2)²*3(x-3)² =(x+2)(x-3)²(2x -6 +3x+6) =5x(x+2)(x-3)² ;
y ' = 0 ⇔ (x+2)x(x-3)² = 0 ⇒ x₁ = - 2 , x₂ =0 , x₃ =3 .
y ' " -" " + " " - " " + "
------------------ (-2) ----------------( 0) ---------------(3) --------------
y ↓ min ↑ max ↓ min ↑
----
8.
y =(x-5)eˣ
y ' = (x-5) '*eˣ +(x-5)*(eˣ) ' =1*eˣ +(x-5)*eˣ =(x-4)*eˣ
y ' = 0 ⇔(x-4)*eˣ = 0 ⇒ x =4 .
y ' " - " " + "
------------------ (4) ---------------
↓ min ↑
Verified answer
',".',".',".',".".".'."".."".