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tanjamichailowa
@tanjamichailowa
June 2021
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12-13 прошу помогите
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m11m
12.
у=
x⁴
-
3x²
+2x
2 2
y' = (1/2) * 4x³ - (3/2) * 2x +2*1=2x³ -3x+2
При х₀=2
у' =2 * 2³ -3*2 +2=16-6+2=12
Ответ: 12
13.
у=2x³ +3x²
y' =2 *3x² +3*2x=6x²+6x=6x(x+1)
y' =0
6x(x+1)=0
x=0 x=-1
+ - +
-------- -1 ------------- 0 -----------------
x=-1 y=2*(-1)³ + 3*(-1)² =-2 +3=1
(-1; 1) - точка максимума
х=0 у=2 * 0³ + 3*0² =0
(0; 0) - точка минимума
Ответ: (-1; 1)
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Answers & Comments
у=x⁴ - 3x² +2x
2 2
y' = (1/2) * 4x³ - (3/2) * 2x +2*1=2x³ -3x+2
При х₀=2
у' =2 * 2³ -3*2 +2=16-6+2=12
Ответ: 12
13.
у=2x³ +3x²
y' =2 *3x² +3*2x=6x²+6x=6x(x+1)
y' =0
6x(x+1)=0
x=0 x=-1
+ - +
-------- -1 ------------- 0 -----------------
x=-1 y=2*(-1)³ + 3*(-1)² =-2 +3=1
(-1; 1) - точка максимума
х=0 у=2 * 0³ + 3*0² =0
(0; 0) - точка минимума
Ответ: (-1; 1)