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erdos7777
@erdos7777
August 2022
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Решите неравенства методом интервала f'(x)<0
a) f(x)=3/x-x^2
b) f(x)=x^4-4,5x^2+2
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sedinalana
Verified answer
F(x)=3/x-x²
f`(x)=-3/x²-2x<0
(-3-2x³)/x²<0
x=-√6/2 x=0
+ _ _
------------------(-√6/2)-------------(0)--------------------
x∈(-√6/2;0) U (0;∞)
f(x)=x^4-4,5x²+2
f`(x)=4x³-9x<0
x(4x²-9)<0
x(2x-3)(2x+3)<0
x=0 x=1,5 x=-1,5
_ + _ +
---------------(-1,5)-------------(0)---------------------(1,5)--------------
x∈(-∞;-1,5) U (0;1,5)
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Answers & Comments
Verified answer
F(x)=3/x-x²f`(x)=-3/x²-2x<0
(-3-2x³)/x²<0
x=-√6/2 x=0
+ _ _
------------------(-√6/2)-------------(0)--------------------
x∈(-√6/2;0) U (0;∞)
f(x)=x^4-4,5x²+2
f`(x)=4x³-9x<0
x(4x²-9)<0
x(2x-3)(2x+3)<0
x=0 x=1,5 x=-1,5
_ + _ +
---------------(-1,5)-------------(0)---------------------(1,5)--------------
x∈(-∞;-1,5) U (0;1,5)