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linamakarovich7
@linamakarovich7
July 2022
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moboqe
1)f(x)=sqrt(3x+1), x0=5
f'(x)=3/(2sqrt(3x+1))
f'(x0)=3/(2sqrt(16))=3/8
2)f(x)=sin^5(x), x0=pi/3
f'(x)=5sin^4(x)*cosx
f'(x0)=5sin^4(pi/3)*cos(pi/3)=5*(sqrt(3)/2)^4*(1/2)=5/2*3^2/2^4=5*9/2^5=45/32
3)y=x^3-6x^2+9x-11
y'=3x^2-12x+9
3x^2-12x+9=0
x^2-4x+3=0
D=16-12=4
x1=3, x2=1
0 votes
Thanks 1
linamakarovich7
Во втором примере pi это что?
moboqe
Это число Пи
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Answers & Comments
f'(x)=3/(2sqrt(3x+1))
f'(x0)=3/(2sqrt(16))=3/8
2)f(x)=sin^5(x), x0=pi/3
f'(x)=5sin^4(x)*cosx
f'(x0)=5sin^4(pi/3)*cos(pi/3)=5*(sqrt(3)/2)^4*(1/2)=5/2*3^2/2^4=5*9/2^5=45/32
3)y=x^3-6x^2+9x-11
y'=3x^2-12x+9
3x^2-12x+9=0
x^2-4x+3=0
D=16-12=4
x1=3, x2=1