Reciprocal Graph - Inverse function

Asked Aug 25 '22 at 12:21

Active Aug 25 '22 at 12:21

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f maps x from (0, positive infinity) to the set of positive real numbers, by $f(x)= 1/x.$ Does the inverse function exist ?

It is injective I believe, so I need to check for surjectivity too. This means the codomain must equal the range. I would say yes because the positive real numbers don't include zero, but the solution says the inverse function doesn't exist as the function is not "onto".

Does this mean they are saying 0 is included in the set of positive real numbers ? Thanks.

asked Aug 25 '22 at 12:21

See here. – SoG Aug 25 '22 at 13:31
The range of $f$ is $(0, \infty) $ . $a\in(0, \infty) $ the preimage of $a$ is $\frac{1}{a}$ as $f(\frac{1}{a}) =a$ – SoG Aug 25 '22 at 13:33
@SouravGhosh So am i correct then ? – Aug 25 '22 at 14:22
1

Yes. See the above link for conformation. – SoG Aug 25 '22 at 14:53

Reciprocal Graph - Inverse function

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