Find the range of the function $f(x)=\sin^{-1}((1+e^x)^{-1})$.
Answer & explanation
Correct answer: option 2
$e^x∈(0,∞)$
$e^x+1∈(1,∞)$
$(e^x+1)^{-1}∈(0,1)$
so $\sin^{-1}(e^x+1)^{-1}∈(0,\frac{π}{2})$
Find the range of the function $f(x)=\sin^{-1}((1+e^x)^{-1})$.
Correct answer: option 2
$e^x∈(0,∞)$
$e^x+1∈(1,∞)$
$(e^x+1)^{-1}∈(0,1)$
so $\sin^{-1}(e^x+1)^{-1}∈(0,\frac{π}{2})$