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Yullwan
@Yullwan
August 2022
2
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1)пользуясь правилами Лопиталя вычислите предел:
lim x->2 (x^3+x-10)/(x^3-3x-2)
2)дана функция f(x)=1/(x-2) найдите f'(x), f"(x), f'"(x)
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Answers & Comments
Noctus
Lim((x³+x-10)/(x³-3x-2))=lim((x³+x-10)'/(x³-3x-2)')=lim((3x²+1)/(3x-3))=
x->2 x->2 x->2
=(3*2²+1)/(3*2²-3)=
13/9
2. f(x)=1/(x-2), f(x)=(x-2)⁻¹
f'(x)
=((x-2)⁻¹)'=-1*(x-2)⁻² *(x-2)'=-1*(x-2)⁻² *1=
-1/((x-2)²)
f''(x)
=((x-2)⁻¹)'=(-(x-2)⁻²)'=-1*(-2)*(x-2)⁻³=2*(x-2)⁻³=
2/((x-2)³)
f'''(x)
=((x-2)⁻¹)'''=(2/(x-2)⁻³)'=2*(-3)*(x-2)⁻⁴=
-6/((x-2)⁴)
0 votes
Thanks 1
kirichekov
Verified answer
1) за f - обозначу знаменатель а за g числительLim f/g = lim x->2 (3x^2+1)/(3x^2-3) = 13/92)f'(x)=-1/(x-2)^2f''(x)=2/(x-2)^3f'''(x)=-6/(x-2)^4
2 votes
Thanks 1
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Answers & Comments
x->2 x->2 x->2
=(3*2²+1)/(3*2²-3)=13/9
2. f(x)=1/(x-2), f(x)=(x-2)⁻¹
f'(x)=((x-2)⁻¹)'=-1*(x-2)⁻² *(x-2)'=-1*(x-2)⁻² *1=-1/((x-2)²)
f''(x)=((x-2)⁻¹)'=(-(x-2)⁻²)'=-1*(-2)*(x-2)⁻³=2*(x-2)⁻³=2/((x-2)³)
f'''(x)=((x-2)⁻¹)'''=(2/(x-2)⁻³)'=2*(-3)*(x-2)⁻⁴=-6/((x-2)⁴)
Verified answer
1) за f - обозначу знаменатель а за g числительLim f/g = lim x->2 (3x^2+1)/(3x^2-3) = 13/92)f'(x)=-1/(x-2)^2f''(x)=2/(x-2)^3f'''(x)=-6/(x-2)^4