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kopylovaulyana
@kopylovaulyana
July 2022
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[tex] \sqrt{-x^2+8x-7} [/tex] Найдите:
1)Область определения функции
2)Промежутки возрастания и убывания функции
3)Наибольшее и наименьшее значение функции на промежутке [3;7]
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sedinalana
Verified answer
1)x²-8x+7≤0
x1+x2=8 U x1*x2=7
x1=1 U x2=7
x∈[1;7]
2)(√(-x²+8x-7))`=(-2x+8)/2√(-x²+8x-7)=0
-2x+8=0
x=4
+ _
[1]--------------------(4)---------------------[7]
возр убыв
y(3)=√(-9+24-7)=√8=2√2
y(4)=√(-16+32-7)=√9=3 наиб
y(7)=√(-49+56-7)=0 наим
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Answers & Comments
Verified answer
1)x²-8x+7≤0x1+x2=8 U x1*x2=7
x1=1 U x2=7
x∈[1;7]
2)(√(-x²+8x-7))`=(-2x+8)/2√(-x²+8x-7)=0
-2x+8=0
x=4
+ _
[1]--------------------(4)---------------------[7]
возр убыв
y(3)=√(-9+24-7)=√8=2√2
y(4)=√(-16+32-7)=√9=3 наиб
y(7)=√(-49+56-7)=0 наим