第1个回答 2022-08-20
过程与结果如图所示
第2个回答 2022-08-20
(1)
f(cosx)
=(sinx)^2/cos2x
=[1-(cosx)^2]/[2(cosx)^2 -1]
f(x)
=(1-x^2)/(2x^2-1)
(2)
f(x)= x/(1+2x)
f(x)+2xf(x) = x
f(x)=x(1-2f(x))
x= f(x)/(1-2f(x))
反函数
f^(-1)(x) = x/(1-2x)
f^(-1)(1) = 1/(1-2)=-1
(3)
lim(x->无穷) [√(x^2+1)-√x^2-1)]
=lim(x->无穷) [(x^2+1)-(x^2-1)]/[√(x^2+1)+√x^2-1)]
=lim(x->无穷) 2/[√(x^2+1)+√x^2-1)]
=0
(4)
f(x)=|x-1|
lim(x->1+) f(x) = lim(x->1+) (x-1) =0
lim(x->1-) f(x) = lim(x->1+) -(x-1) =0
=>
lim(x->1) f(x)=0
x=1, f(x) 连续
ans :D
f(cosx)
=(sinx)^2/cos2x
=[1-(cosx)^2]/[2(cosx)^2 -1]
f(x)
=(1-x^2)/(2x^2-1)
(2)
f(x)= x/(1+2x)
f(x)+2xf(x) = x
f(x)=x(1-2f(x))
x= f(x)/(1-2f(x))
反函数
f^(-1)(x) = x/(1-2x)
f^(-1)(1) = 1/(1-2)=-1
(3)
lim(x->无穷) [√(x^2+1)-√x^2-1)]
=lim(x->无穷) [(x^2+1)-(x^2-1)]/[√(x^2+1)+√x^2-1)]
=lim(x->无穷) 2/[√(x^2+1)+√x^2-1)]
=0
(4)
f(x)=|x-1|
lim(x->1+) f(x) = lim(x->1+) (x-1) =0
lim(x->1-) f(x) = lim(x->1+) -(x-1) =0
=>
lim(x->1) f(x)=0
x=1, f(x) 连续
ans :D
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