Question 895e9

Sep 4, 2017

$\text{Dom} \left(f\right) = u \in \mathbb{R} , u \ne - 1$
$\text{Ran} \left(f\right) = y \in \mathbb{R} , y \ne 1$
The function y-=f(u)=1"/"(1+1"/"u)# can be rewritten as $f \left(u\right) = u \text{/} \left(u + 1\right)$.
From the rewritten form, it's clear that $u \ne - 1$ as this would cause the denominator to be equal to zero. From the original form, we can see that a horizontal asymptote will occur at $y = 1$. This is because as $u \rightarrow \infty$, $y \rightarrow 1$.