Question

求救 函数题已知函数f(x)=1/(√(x^2-4)) x∈(-∞,-2)1.求f(x)的反函数g(x)2.设a(1)=

Answer 1

x∈(-∞,-2)

=>

x^2-4∈(0,+∞)

=>

√(x^2-4)∈(0,+∞)

=>

y=f(x)=1/(√(x^2-4)) ∈(0,+∞)

=>

反函数定义域为(0,+∞)

y=1/(√(x^2-4))

=>

√(x^2-4)=1/y

=>

x^2-4=1/y^2

=>

x=-√(1/y^2+4)

=>

反函数

g(x)=-√(1/x^2+4) (x∈(0,+∞))

1/a(n+1)=-g(a(n))

=>

1/(a(n+1))=√(1/(a(n))^2+4)

=>

1/(a(n+1))^2=1/(a(n))^2+4

=>

｛1/a(n)^2｝是等差数列

=>

1/(a(n))^2

=4(n-1)+1/(a1)^2

=4(n-1)+1

=4n-3

=>

(a(n))^2=1/(4n-3)

=>

a(n)=√(1/(4n-3))

b(n)=S(n+1)-S(n)

=>

b(n)

=(a(n+1))^2

=1/(4n+1)

欲使

1/(4n+1)5

=>

min[m]=6