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ghagha0
@ghagha0
September 2021
1
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найдите точку минимума функции
[tex] -\frac{x}{ x^{2} +225} [/tex]
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sedinalana
Verified answer
F(x)=-x/(x²+225)
f`(x)=(-x²-225+2x²)/(x²+225)²=(x²-225)/(x²+225)²=0
x²-225=0
x²=225
x=-15 x=15
+ _ +
-------------(-15)------------(15)---------------
min
minf(15)=-15/(225+225)=-15/450=-1/30
16 votes
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Answers & Comments
Verified answer
F(x)=-x/(x²+225)f`(x)=(-x²-225+2x²)/(x²+225)²=(x²-225)/(x²+225)²=0
x²-225=0
x²=225
x=-15 x=15
+ _ +
-------------(-15)------------(15)---------------
min
minf(15)=-15/(225+225)=-15/450=-1/30