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lacosanostraone
@lacosanostraone
July 2022
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Найдите наименьшее значение функции y=x²-8x+7
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sangers1959
Verified answer
Y=x²-8x+7
y`=(x²-8x+7)`=2x-8=0
2x-8=0 |÷2
x-4=0
x=4 ⇒
ymin=4²-8*4+7=16-32+7=-9.
Ответ: ymin=-9.
2 votes
Thanks 4
giglesshow
y=x²-8x+7
y`=(x²-8x+7)`=2x-8=0
2x-8=0 |÷2
x-4=0
x=4 ⇒
ymin=4²-8*4+7=16-32+7=-9.
Ответ: ymin=-9.
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Answers & Comments
Verified answer
Y=x²-8x+7y`=(x²-8x+7)`=2x-8=0
2x-8=0 |÷2
x-4=0
x=4 ⇒
ymin=4²-8*4+7=16-32+7=-9.
Ответ: ymin=-9.
y`=(x²-8x+7)`=2x-8=0
2x-8=0 |÷2
x-4=0
x=4 ⇒
ymin=4²-8*4+7=16-32+7=-9.
Ответ: ymin=-9.