Pretty self explanatory. I'm trying to find the range of f(x)=arctan(1+1x)
but in all honesty, I'm not really sure how to proceed. I feel like there is something very silly and obvious I'm missing. I can find the domain fairly easily but how do I go about finding the range without taking its inverse? Is that even possible?
I have the same problem with:
f(x)=ex+√x2+1
In general, how would I go about finding the range? I know I can just take the inverse and find the domain of that, but is that the only way? Thanks!
Answer
At first we need to show the range for f(x)=1+1x which is of course (−∞,∞)∖{1}, indeed
y=1+1x⟺x=1y−1
therefore we can reach any value but not y=1.
For the second one we have that x+√x2+1>0 and
limx→∞x+√x2+1=∞
limx→−∞x+√x2+1=limu→∞−u+√u2+1=limu→∞−u2+u2+1u+√u2+1=0
In order to avoid limits note that
y=x+√x2+1⟺(y−x)2=x2+1⟺y2−2xy+x2=x2+1⟺x=y2−12y
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