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sementuzov1
@sementuzov1
July 2022
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4
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Решите неравенство (cos2x+3tgпи/8)'>=2sinx
Найдите производную функции f(x)=(3x+2)^3*(2x-1)^4
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Godnessgirl
1) 2sin2x≥2sinx
sin2x≥sinx
2sinx cosx≥sinx
2sinx cosx - sinx≥0
sinx(2cosx-1)≥0
Делим га на две системы:
1. sinx≥0; [2πn; π+2πn]
2cosx-1≥0; [-π/3+2πn; π/3+2πn]
Ответ: хє[2πn;π/3+2πn]
2. sinx≤0; [π+2πn; 2πn]
2cosx-1≤0; [π/3+2πn; 5π/3+2πn]
Ответ: хє[π+2πn; 5π/3+2πn].
2) f(x)=(3x+2)^3*(2x-1)^4
f'(x)= 3(3x+2)²*3(2x-1)⁴+4(2x-1)³*2(3x+2)³= 9(3x+2)²(2x-1)⁴+8(3x+2)³(2x-1)³= (3x+2)²(2x-1)³(9(2x-1)+8(3x+2))=(3x+2)²(2x-1)³(42x+7)=7(3x+2)²(2x-1)³(6x+1).
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Answers & Comments
sin2x≥sinx
2sinx cosx≥sinx
2sinx cosx - sinx≥0
sinx(2cosx-1)≥0
Делим га на две системы:
1. sinx≥0; [2πn; π+2πn]
2cosx-1≥0; [-π/3+2πn; π/3+2πn]
Ответ: хє[2πn;π/3+2πn]
2. sinx≤0; [π+2πn; 2πn]
2cosx-1≤0; [π/3+2πn; 5π/3+2πn]
Ответ: хє[π+2πn; 5π/3+2πn].
2) f(x)=(3x+2)^3*(2x-1)^4
f'(x)= 3(3x+2)²*3(2x-1)⁴+4(2x-1)³*2(3x+2)³= 9(3x+2)²(2x-1)⁴+8(3x+2)³(2x-1)³= (3x+2)²(2x-1)³(9(2x-1)+8(3x+2))=(3x+2)²(2x-1)³(42x+7)=7(3x+2)²(2x-1)³(6x+1).