Silly doubt regarding onto function












1












$begingroup$



Let function $f:(0,infty)to(0,infty)$ be defined as $f(x)=|1-frac{1}{x}|$. Then is it onto function?




My doubt is that here the codomain doesn't include $0$ but here in the function $0$ is there on $x=1$. So, will it be an onto function? (confusion is created due to this point $(1,0) $ ).



Thanks for clearing my doubt.






algebra-precalculus functions self-learning







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$endgroup$












  • $begingroup$
    You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
    $endgroup$
    – lulu
    Jan 17 at 13:36










  • $begingroup$
    @lulu No, actually it has been given in the test called for JEE Main.
    $endgroup$
    – jayant98
    Jan 17 at 13:38










  • $begingroup$
    It could still be a typo. It's very hard to eliminate typos.
    $endgroup$
    – lulu
    Jan 17 at 13:40
















1












$begingroup$



Let function $f:(0,infty)to(0,infty)$ be defined as $f(x)=|1-frac{1}{x}|$. Then is it onto function?




My doubt is that here the codomain doesn't include $0$ but here in the function $0$ is there on $x=1$. So, will it be an onto function? (confusion is created due to this point $(1,0) $ ).



Thanks for clearing my doubt.






algebra-precalculus functions self-learning







share|cite|improve this question











$endgroup$












  • $begingroup$
    You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
    $endgroup$
    – lulu
    Jan 17 at 13:36










  • $begingroup$
    @lulu No, actually it has been given in the test called for JEE Main.
    $endgroup$
    – jayant98
    Jan 17 at 13:38










  • $begingroup$
    It could still be a typo. It's very hard to eliminate typos.
    $endgroup$
    – lulu
    Jan 17 at 13:40














1












1








1





$begingroup$



Let function $f:(0,infty)to(0,infty)$ be defined as $f(x)=|1-frac{1}{x}|$. Then is it onto function?




My doubt is that here the codomain doesn't include $0$ but here in the function $0$ is there on $x=1$. So, will it be an onto function? (confusion is created due to this point $(1,0) $ ).



Thanks for clearing my doubt.






algebra-precalculus functions self-learning







share|cite|improve this question











$endgroup$





Let function $f:(0,infty)to(0,infty)$ be defined as $f(x)=|1-frac{1}{x}|$. Then is it onto function?




My doubt is that here the codomain doesn't include $0$ but here in the function $0$ is there on $x=1$. So, will it be an onto function? (confusion is created due to this point $(1,0) $ ).



Thanks for clearing my doubt.






algebra-precalculus functions self-learning




algebra-precalculus functions self-learning






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 13:35









Thomas Shelby

4,7382727




4,7382727










asked Jan 17 at 13:27









jayant98jayant98

649318




649318












  • $begingroup$
    You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
    $endgroup$
    – lulu
    Jan 17 at 13:36










  • $begingroup$
    @lulu No, actually it has been given in the test called for JEE Main.
    $endgroup$
    – jayant98
    Jan 17 at 13:38










  • $begingroup$
    It could still be a typo. It's very hard to eliminate typos.
    $endgroup$
    – lulu
    Jan 17 at 13:40


















  • $begingroup$
    You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
    $endgroup$
    – lulu
    Jan 17 at 13:36










  • $begingroup$
    @lulu No, actually it has been given in the test called for JEE Main.
    $endgroup$
    – jayant98
    Jan 17 at 13:38










  • $begingroup$
    It could still be a typo. It's very hard to eliminate typos.
    $endgroup$
    – lulu
    Jan 17 at 13:40
















$begingroup$
You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
$endgroup$
– lulu
Jan 17 at 13:36




$begingroup$
You are correct, but I expect the question was intended to refer to $f$ as mapping to $[0,infty)$. With that change it makes sense to ask if it is surjective or not.
$endgroup$
– lulu
Jan 17 at 13:36












$begingroup$
@lulu No, actually it has been given in the test called for JEE Main.
$endgroup$
– jayant98
Jan 17 at 13:38




$begingroup$
@lulu No, actually it has been given in the test called for JEE Main.
$endgroup$
– jayant98
Jan 17 at 13:38












$begingroup$
It could still be a typo. It's very hard to eliminate typos.
$endgroup$
– lulu
Jan 17 at 13:40




$begingroup$
It could still be a typo. It's very hard to eliminate typos.
$endgroup$
– lulu
Jan 17 at 13:40










1 Answer
1






active

oldest

votes


















1












$begingroup$

You are right, $f$ is not a well defined function and the reason has been stated by you. $f(1)$ is not in the codomain.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So it is not onto function.
    $endgroup$
    – jayant98
    Jan 17 at 13:30






  • 1




    $begingroup$
    it is not even a function.
    $endgroup$
    – Siong Thye Goh
    Jan 17 at 13:30










  • $begingroup$
    Okay. I understood. Thank you for helping me.
    $endgroup$
    – jayant98
    Jan 17 at 13:31






  • 1




    $begingroup$
    It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
    $endgroup$
    – Jochen
    Jan 17 at 13:32












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You are right, $f$ is not a well defined function and the reason has been stated by you. $f(1)$ is not in the codomain.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So it is not onto function.
    $endgroup$
    – jayant98
    Jan 17 at 13:30






  • 1




    $begingroup$
    it is not even a function.
    $endgroup$
    – Siong Thye Goh
    Jan 17 at 13:30










  • $begingroup$
    Okay. I understood. Thank you for helping me.
    $endgroup$
    – jayant98
    Jan 17 at 13:31






  • 1




    $begingroup$
    It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
    $endgroup$
    – Jochen
    Jan 17 at 13:32
















1












$begingroup$

You are right, $f$ is not a well defined function and the reason has been stated by you. $f(1)$ is not in the codomain.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    So it is not onto function.
    $endgroup$
    – jayant98
    Jan 17 at 13:30






  • 1




    $begingroup$
    it is not even a function.
    $endgroup$
    – Siong Thye Goh
    Jan 17 at 13:30










  • $begingroup$
    Okay. I understood. Thank you for helping me.
    $endgroup$
    – jayant98
    Jan 17 at 13:31






  • 1




    $begingroup$
    It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
    $endgroup$
    – Jochen
    Jan 17 at 13:32














1












1








1





$begingroup$

You are right, $f$ is not a well defined function and the reason has been stated by you. $f(1)$ is not in the codomain.






share|cite|improve this answer









$endgroup$



You are right, $f$ is not a well defined function and the reason has been stated by you. $f(1)$ is not in the codomain.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 13:29









Siong Thye GohSiong Thye Goh

104k1468120




104k1468120












  • $begingroup$
    So it is not onto function.
    $endgroup$
    – jayant98
    Jan 17 at 13:30






  • 1




    $begingroup$
    it is not even a function.
    $endgroup$
    – Siong Thye Goh
    Jan 17 at 13:30










  • $begingroup$
    Okay. I understood. Thank you for helping me.
    $endgroup$
    – jayant98
    Jan 17 at 13:31






  • 1




    $begingroup$
    It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
    $endgroup$
    – Jochen
    Jan 17 at 13:32


















  • $begingroup$
    So it is not onto function.
    $endgroup$
    – jayant98
    Jan 17 at 13:30






  • 1




    $begingroup$
    it is not even a function.
    $endgroup$
    – Siong Thye Goh
    Jan 17 at 13:30










  • $begingroup$
    Okay. I understood. Thank you for helping me.
    $endgroup$
    – jayant98
    Jan 17 at 13:31






  • 1




    $begingroup$
    It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
    $endgroup$
    – Jochen
    Jan 17 at 13:32
















$begingroup$
So it is not onto function.
$endgroup$
– jayant98
Jan 17 at 13:30




$begingroup$
So it is not onto function.
$endgroup$
– jayant98
Jan 17 at 13:30




1




1




$begingroup$
it is not even a function.
$endgroup$
– Siong Thye Goh
Jan 17 at 13:30




$begingroup$
it is not even a function.
$endgroup$
– Siong Thye Goh
Jan 17 at 13:30












$begingroup$
Okay. I understood. Thank you for helping me.
$endgroup$
– jayant98
Jan 17 at 13:31




$begingroup$
Okay. I understood. Thank you for helping me.
$endgroup$
– jayant98
Jan 17 at 13:31




1




1




$begingroup$
It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
$endgroup$
– Jochen
Jan 17 at 13:32




$begingroup$
It is not a (correctly defined) function, hence it is not an onto function. If you change the codomain to $[0,infty)$ it becomes an onto function (either you solve $f(x)=y$ for each $yin [0,infty)$ explicitely or you use the intermediate value theorem).
$endgroup$
– Jochen
Jan 17 at 13:32


















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