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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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&#039;t have an interval [a,b] with a,b   ∈       R  . I know that I can consider the limit, but we didn&#039;t proved the fact, that I can use the intermediate value theorem for limits.

Simeon Hester · Accepted Answer

First show that       |    f      |    &amp;lt;  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  &amp;lt;  0 separately.

trkalo84 · Accepted Answer

|    f    (    x    )    |    =      |          x                        |                x                  |                +        1              |    =                    |            x              |                            |            x              |            +      1        &amp;lt;  1thus as   f  (  x  ) is continuos and lim as   x  →  ±  ∞  =  ±  1  −  1  &amp;lt;  f  (  x  )  &amp;lt;  1

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