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staffanist
@staffanist
July 2022
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найдите наибольшее и наименьшее значение функции у=2х²+3х+5 на отрезке (-3;1)
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BParCuCTeMbll337
1) y' = 4x +3
4x + 3 = 0
4x = -3
x = -3/4
2) а) х = -3
у = 2*(-3)² + 3*(-3) +5 = 18 - 9 +5 = 14
б) х = 1
у = 2*1² +3*1 +5 = 10
в) х = -3/4
у = 2*(-3/4)² + 3*(-3/4) +5 = 9/8 - 9/4 +5= 31/8 = 3,875
Ответ: уmax = 14
y min = 3,875
1 votes
Thanks 1
sharadi
Verified answer
Y=2x^2+3x+5 (-3;1)
y'=4x+3
4x+3=0
x=-3/4
y(-3)=2*(-3)^2+3*(-3)+5
y(-3)=14
y(-3/4)=2*(-3/4)^2+3*(-3/4)+5
y(-3/4)=3,3
y(1)=2*1^2+3*1+5=10
y(min)=y(-3/4) y(max)=y(-3)
0 votes
Thanks 1
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Answers & Comments
4x + 3 = 0
4x = -3
x = -3/4
2) а) х = -3
у = 2*(-3)² + 3*(-3) +5 = 18 - 9 +5 = 14
б) х = 1
у = 2*1² +3*1 +5 = 10
в) х = -3/4
у = 2*(-3/4)² + 3*(-3/4) +5 = 9/8 - 9/4 +5= 31/8 = 3,875
Ответ: уmax = 14
y min = 3,875
Verified answer
Y=2x^2+3x+5 (-3;1)y'=4x+3
4x+3=0
x=-3/4
y(-3)=2*(-3)^2+3*(-3)+5
y(-3)=14
y(-3/4)=2*(-3/4)^2+3*(-3/4)+5
y(-3/4)=3,3
y(1)=2*1^2+3*1+5=10
y(min)=y(-3/4) y(max)=y(-3)