ƒ (x) = sin(x) + cos(2x)
ƒ' (x) =cos (x) -2sin(2x)
cos (x) -2sin(2x)=0
cos (x) -4sin(x)cos (x)=0
cos (x)(1-4sin(x))=0
cos (x)=0 или 4sin(x)=1
х=π/2+πn или [tex]x=(-1)^{n} arccsin\frac{1}{4} +\pi n[/tex]
f'(x) + - + - +
______(arcsin(1/4))___(π/2)____(π-arcsin(1/4))_____(3π/2)_____
f(x) ↑ ↓ ↑ ↓ ↑
Определим знак производной (arcsin(1/4) ;π/2)
при х=π/6∈(arcsin(1/4) ;
[tex]f' (\frac{\pi }{6} ) =cos (\frac{\pi }{6} ) -2sin(2*\frac{\pi }{6} )=cos (\frac{\pi }{6} ) -2sin(\frac{\pi }{3} )=\frac{\sqrt{3} }{2} -\frac{2}{1} *\frac{\sqrt{3} }{2} =-\frac{\sqrt{3} }{2} < 0[/tex]
x=arcsin(1/4)+ 2πn , max
x=π/2+2πn, min
x=π-arcsin(1/4)+ 2πn , max
x=3π/2+2πn, min
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Answers & Comments
ƒ (x) = sin(x) + cos(2x)
ƒ' (x) =cos (x) -2sin(2x)
cos (x) -2sin(2x)=0
cos (x) -4sin(x)cos (x)=0
cos (x)(1-4sin(x))=0
cos (x)=0 или 4sin(x)=1
х=π/2+πn или [tex]x=(-1)^{n} arccsin\frac{1}{4} +\pi n[/tex]
f'(x) + - + - +
______(arcsin(1/4))___(π/2)____(π-arcsin(1/4))_____(3π/2)_____
f(x) ↑ ↓ ↑ ↓ ↑
Определим знак производной (arcsin(1/4) ;π/2)
при х=π/6∈(arcsin(1/4) ;
[tex]f' (\frac{\pi }{6} ) =cos (\frac{\pi }{6} ) -2sin(2*\frac{\pi }{6} )=cos (\frac{\pi }{6} ) -2sin(\frac{\pi }{3} )=\frac{\sqrt{3} }{2} -\frac{2}{1} *\frac{\sqrt{3} }{2} =-\frac{\sqrt{3} }{2} < 0[/tex]
x=arcsin(1/4)+ 2πn , max
x=π/2+2πn, min
x=π-arcsin(1/4)+ 2πn , max
x=3π/2+2πn, min