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8AKULA8
@8AKULA8
March 2022
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40 баллов. Срочно нужно
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mikael2
1. 2√x+3/x² y'=1/√x-3*2/x³=1/√x-6/x³ x=1 y'=1-6=-5
2. h(x)=cos2x h'(x)=-2sin2x h'(-π/3)=-h'(π/3)=2sin2π/3=2*√3/2=√3
-2sin2x=1 sin2x=-1/2 2x=-(1)^n*π/6+πn x= -(1)^n*π/12+πn/2 n∈z
3. f(x)=2x+7/x-4 f'(x)=1/u²[u'v-v'u] u=2x+7 u'=2 v=x-4 v'=1
f'(x)=1/(x-4)²[2*x-8-2x-7]=-15/(x-4)² f'(5)=-15
-15/(x-4)² <0 x∈R x≠4
4. f(x)=cos²x f'(x)=-2cosx*sinx=-sin2x
-sin2x=1 sin2x=-1/2 2x=-(-1)^n*π/6+πn
x=(-1)^(n+1)*π/12+πn/2
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Answers & Comments
2. h(x)=cos2x h'(x)=-2sin2x h'(-π/3)=-h'(π/3)=2sin2π/3=2*√3/2=√3
-2sin2x=1 sin2x=-1/2 2x=-(1)^n*π/6+πn x= -(1)^n*π/12+πn/2 n∈z
3. f(x)=2x+7/x-4 f'(x)=1/u²[u'v-v'u] u=2x+7 u'=2 v=x-4 v'=1
f'(x)=1/(x-4)²[2*x-8-2x-7]=-15/(x-4)² f'(5)=-15
-15/(x-4)² <0 x∈R x≠4
4. f(x)=cos²x f'(x)=-2cosx*sinx=-sin2x
-sin2x=1 sin2x=-1/2 2x=-(-1)^n*π/6+πn
x=(-1)^(n+1)*π/12+πn/2