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)=

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