Consider the function: f:R->R f(x)=x/(|x|+1) I already showed that this function is continuous and injective. Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b in R. I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits.

saucletbh 2022-09-17 Answered
Consider the function:
f : R R
f ( x ) = x | x | + 1
I already showed that this function is continuous and injective. Maybe this could help? Maybe we can use the intermediate value theorem? The problem is that I don't have an interval [a,b] with a,b R . I know that I can consider the limit, but we didn't proved the fact, that I can use the intermediate value theorem for limits.
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Answers (2)

Simeon Hester
Answered 2022-09-18 Author has 6 answers
First show that | f | < 1 (this should be pretty easy). Then take any y ( 1 , 1 ) and show that you can explicitly solve f ( x ) = y for x, thereby showing that the image is ( 1 , 1 ) and nothing else. You might find it helpful to do y 0 and y < 0 separately.
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trkalo84
Answered 2022-09-19 Author has 2 answers
| f ( x ) | = | x | x | + 1 | = | x | | x | + 1 < 1
thus as f ( x ) is continuos and lim as x ± = ± 1
1 < f ( x ) < 1
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