How do I determine the inverse function?

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I have this exercise;



For each of the following functions, determine the inverse function.



Here, $mathbbR_geq 0$ denotes the set of all non-negative reals:



$$f : mathbbR_geq 0tomathbbR, xmapsto sqrtx$$



But I really don't know where or how to start, could anyone provide some guidance on how to get an inverse function? maybe show some steps? Thank you so much







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    up vote
    0
    down vote

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    I have this exercise;



    For each of the following functions, determine the inverse function.



    Here, $mathbbR_geq 0$ denotes the set of all non-negative reals:



    $$f : mathbbR_geq 0tomathbbR, xmapsto sqrtx$$



    But I really don't know where or how to start, could anyone provide some guidance on how to get an inverse function? maybe show some steps? Thank you so much







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have this exercise;



      For each of the following functions, determine the inverse function.



      Here, $mathbbR_geq 0$ denotes the set of all non-negative reals:



      $$f : mathbbR_geq 0tomathbbR, xmapsto sqrtx$$



      But I really don't know where or how to start, could anyone provide some guidance on how to get an inverse function? maybe show some steps? Thank you so much







      share|cite|improve this question














      I have this exercise;



      For each of the following functions, determine the inverse function.



      Here, $mathbbR_geq 0$ denotes the set of all non-negative reals:



      $$f : mathbbR_geq 0tomathbbR, xmapsto sqrtx$$



      But I really don't know where or how to start, could anyone provide some guidance on how to get an inverse function? maybe show some steps? Thank you so much









      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Aug 22 at 11:39









      Andrés E. Caicedo

      63.3k7152237




      63.3k7152237










      asked Aug 22 at 10:54









      ValentineJ

      51




      51




















          4 Answers
          4






          active

          oldest

          votes

















          up vote
          2
          down vote













          In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.



          For your example, observe that $y = sqrtx$ so that $y^2 = x$. Thus your inverse function is $f^-1(x) = x^2$.



          It remains to determine the domain and range. We first note that $f : mathbbR_geq 0 to mathbbR$ has the domain $mathbbR_geq 0$ and the range $mathbbR$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $mathbbR_geq 0$ (can you see why?). Thus, the domain of $f^-1$ will be this set, the image of $f$, and the range of $f^-1$ will be the domain of $f$ so that $f^-1 : mathbbR_geq 0 to mathbbR_geq 0$. Hence the inverse function is
          $$
          f^-1 : mathbbR_geq 0 to mathbbR_geq 0,quad f^-1(x) = x^2.
          $$



          It is important that the domain of $f^-1$ is the image of $f$ otherwise, as drhab points out, one may have something like
          $$
          -1stackrelf^-1to1stackrelfto1neq-1,
          $$
          which is not what we want since we should have $fcirc f^-1 = f^-1circ f = textIdentity map$.






          share|cite|improve this answer






















          • Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
            – drhab
            Aug 22 at 12:13











          • @drhab Ah yes, let me fix that!
            – Bill Wallis
            Aug 22 at 12:15

















          up vote
          1
          down vote













          We have $f(R)=R$ and $f$ is injective.



          $y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.






          share|cite|improve this answer




















          • What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
            – drhab
            Aug 22 at 12:14










          • In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
            – Fred
            Aug 22 at 13:05

















          up vote
          1
          down vote













          You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y

          $$y=sqrtx$$
          $$y^2=x$$
          $$x=y^2$$
          $$f^-1(x)=x^2$$






          share|cite|improve this answer




















          • Sorry for being stupid but does "f−1" denote that it's an inverse function?
            – ValentineJ
            Aug 22 at 11:03










          • yes $f^-1(x)$ is the inverse function of $f(x)$
            – Deepesh Meena
            Aug 22 at 11:03











          • awesome, great explanation thank you
            – ValentineJ
            Aug 22 at 11:04

















          up vote
          1
          down vote













          The function $f:mathbb R_geq0tomathbb R$ prescribed by $xmapstosqrt x$ has no inverse.



          This because it is not surjective.



          The function $g:mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapstosqrt x$ is bijective, hence has an inverse.



          It is the function $mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapsto x^2$.



          For finding this function see the other answers.






          share|cite|improve this answer




















          • At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
            – drhab
            Aug 23 at 17:12











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          4 Answers
          4






          active

          oldest

          votes








          4 Answers
          4






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote













          In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.



          For your example, observe that $y = sqrtx$ so that $y^2 = x$. Thus your inverse function is $f^-1(x) = x^2$.



          It remains to determine the domain and range. We first note that $f : mathbbR_geq 0 to mathbbR$ has the domain $mathbbR_geq 0$ and the range $mathbbR$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $mathbbR_geq 0$ (can you see why?). Thus, the domain of $f^-1$ will be this set, the image of $f$, and the range of $f^-1$ will be the domain of $f$ so that $f^-1 : mathbbR_geq 0 to mathbbR_geq 0$. Hence the inverse function is
          $$
          f^-1 : mathbbR_geq 0 to mathbbR_geq 0,quad f^-1(x) = x^2.
          $$



          It is important that the domain of $f^-1$ is the image of $f$ otherwise, as drhab points out, one may have something like
          $$
          -1stackrelf^-1to1stackrelfto1neq-1,
          $$
          which is not what we want since we should have $fcirc f^-1 = f^-1circ f = textIdentity map$.






          share|cite|improve this answer






















          • Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
            – drhab
            Aug 22 at 12:13











          • @drhab Ah yes, let me fix that!
            – Bill Wallis
            Aug 22 at 12:15














          up vote
          2
          down vote













          In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.



          For your example, observe that $y = sqrtx$ so that $y^2 = x$. Thus your inverse function is $f^-1(x) = x^2$.



          It remains to determine the domain and range. We first note that $f : mathbbR_geq 0 to mathbbR$ has the domain $mathbbR_geq 0$ and the range $mathbbR$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $mathbbR_geq 0$ (can you see why?). Thus, the domain of $f^-1$ will be this set, the image of $f$, and the range of $f^-1$ will be the domain of $f$ so that $f^-1 : mathbbR_geq 0 to mathbbR_geq 0$. Hence the inverse function is
          $$
          f^-1 : mathbbR_geq 0 to mathbbR_geq 0,quad f^-1(x) = x^2.
          $$



          It is important that the domain of $f^-1$ is the image of $f$ otherwise, as drhab points out, one may have something like
          $$
          -1stackrelf^-1to1stackrelfto1neq-1,
          $$
          which is not what we want since we should have $fcirc f^-1 = f^-1circ f = textIdentity map$.






          share|cite|improve this answer






















          • Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
            – drhab
            Aug 22 at 12:13











          • @drhab Ah yes, let me fix that!
            – Bill Wallis
            Aug 22 at 12:15












          up vote
          2
          down vote










          up vote
          2
          down vote









          In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.



          For your example, observe that $y = sqrtx$ so that $y^2 = x$. Thus your inverse function is $f^-1(x) = x^2$.



          It remains to determine the domain and range. We first note that $f : mathbbR_geq 0 to mathbbR$ has the domain $mathbbR_geq 0$ and the range $mathbbR$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $mathbbR_geq 0$ (can you see why?). Thus, the domain of $f^-1$ will be this set, the image of $f$, and the range of $f^-1$ will be the domain of $f$ so that $f^-1 : mathbbR_geq 0 to mathbbR_geq 0$. Hence the inverse function is
          $$
          f^-1 : mathbbR_geq 0 to mathbbR_geq 0,quad f^-1(x) = x^2.
          $$



          It is important that the domain of $f^-1$ is the image of $f$ otherwise, as drhab points out, one may have something like
          $$
          -1stackrelf^-1to1stackrelfto1neq-1,
          $$
          which is not what we want since we should have $fcirc f^-1 = f^-1circ f = textIdentity map$.






          share|cite|improve this answer














          In general, if you have a function $y = f(x)$ and you want to find the inverse, then you want to rearrange $y = f(x)$ so that $x$ is a function of $y$.



          For your example, observe that $y = sqrtx$ so that $y^2 = x$. Thus your inverse function is $f^-1(x) = x^2$.



          It remains to determine the domain and range. We first note that $f : mathbbR_geq 0 to mathbbR$ has the domain $mathbbR_geq 0$ and the range $mathbbR$. However, the image of $f$ is the set of all elements in the range that are mapped to by $f$ from something in the domain. In this case, the image of $f$ is $mathbbR_geq 0$ (can you see why?). Thus, the domain of $f^-1$ will be this set, the image of $f$, and the range of $f^-1$ will be the domain of $f$ so that $f^-1 : mathbbR_geq 0 to mathbbR_geq 0$. Hence the inverse function is
          $$
          f^-1 : mathbbR_geq 0 to mathbbR_geq 0,quad f^-1(x) = x^2.
          $$



          It is important that the domain of $f^-1$ is the image of $f$ otherwise, as drhab points out, one may have something like
          $$
          -1stackrelf^-1to1stackrelfto1neq-1,
          $$
          which is not what we want since we should have $fcirc f^-1 = f^-1circ f = textIdentity map$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 22 at 12:19

























          answered Aug 22 at 10:59









          Bill Wallis

          2,2361826




          2,2361826











          • Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
            – drhab
            Aug 22 at 12:13











          • @drhab Ah yes, let me fix that!
            – Bill Wallis
            Aug 22 at 12:15
















          • Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
            – drhab
            Aug 22 at 12:13











          • @drhab Ah yes, let me fix that!
            – Bill Wallis
            Aug 22 at 12:15















          Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
          – drhab
          Aug 22 at 12:13





          Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
          – drhab
          Aug 22 at 12:13













          @drhab Ah yes, let me fix that!
          – Bill Wallis
          Aug 22 at 12:15




          @drhab Ah yes, let me fix that!
          – Bill Wallis
          Aug 22 at 12:15










          up vote
          1
          down vote













          We have $f(R)=R$ and $f$ is injective.



          $y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.






          share|cite|improve this answer




















          • What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
            – drhab
            Aug 22 at 12:14










          • In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
            – Fred
            Aug 22 at 13:05














          up vote
          1
          down vote













          We have $f(R)=R$ and $f$ is injective.



          $y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.






          share|cite|improve this answer




















          • What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
            – drhab
            Aug 22 at 12:14










          • In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
            – Fred
            Aug 22 at 13:05












          up vote
          1
          down vote










          up vote
          1
          down vote









          We have $f(R)=R$ and $f$ is injective.



          $y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.






          share|cite|improve this answer












          We have $f(R)=R$ and $f$ is injective.



          $y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 22 at 10:59









          Fred

          38.2k1238




          38.2k1238











          • What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
            – drhab
            Aug 22 at 12:14










          • In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
            – Fred
            Aug 22 at 13:05
















          • What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
            – drhab
            Aug 22 at 12:14










          • In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
            – Fred
            Aug 22 at 13:05















          What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
          – drhab
          Aug 22 at 12:14




          What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
          – drhab
          Aug 22 at 12:14












          In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
          – Fred
          Aug 22 at 13:05




          In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
          – Fred
          Aug 22 at 13:05










          up vote
          1
          down vote













          You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y

          $$y=sqrtx$$
          $$y^2=x$$
          $$x=y^2$$
          $$f^-1(x)=x^2$$






          share|cite|improve this answer




















          • Sorry for being stupid but does "f−1" denote that it's an inverse function?
            – ValentineJ
            Aug 22 at 11:03










          • yes $f^-1(x)$ is the inverse function of $f(x)$
            – Deepesh Meena
            Aug 22 at 11:03











          • awesome, great explanation thank you
            – ValentineJ
            Aug 22 at 11:04














          up vote
          1
          down vote













          You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y

          $$y=sqrtx$$
          $$y^2=x$$
          $$x=y^2$$
          $$f^-1(x)=x^2$$






          share|cite|improve this answer




















          • Sorry for being stupid but does "f−1" denote that it's an inverse function?
            – ValentineJ
            Aug 22 at 11:03










          • yes $f^-1(x)$ is the inverse function of $f(x)$
            – Deepesh Meena
            Aug 22 at 11:03











          • awesome, great explanation thank you
            – ValentineJ
            Aug 22 at 11:04












          up vote
          1
          down vote










          up vote
          1
          down vote









          You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y

          $$y=sqrtx$$
          $$y^2=x$$
          $$x=y^2$$
          $$f^-1(x)=x^2$$






          share|cite|improve this answer












          You can solve the question in the follwoing manner swap x and y then you will get a function $x=f(y)$ in terms of $y$ again swap x and y

          $$y=sqrtx$$
          $$y^2=x$$
          $$x=y^2$$
          $$f^-1(x)=x^2$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 22 at 11:01









          Deepesh Meena

          2,635719




          2,635719











          • Sorry for being stupid but does "f−1" denote that it's an inverse function?
            – ValentineJ
            Aug 22 at 11:03










          • yes $f^-1(x)$ is the inverse function of $f(x)$
            – Deepesh Meena
            Aug 22 at 11:03











          • awesome, great explanation thank you
            – ValentineJ
            Aug 22 at 11:04
















          • Sorry for being stupid but does "f−1" denote that it's an inverse function?
            – ValentineJ
            Aug 22 at 11:03










          • yes $f^-1(x)$ is the inverse function of $f(x)$
            – Deepesh Meena
            Aug 22 at 11:03











          • awesome, great explanation thank you
            – ValentineJ
            Aug 22 at 11:04















          Sorry for being stupid but does "f−1" denote that it's an inverse function?
          – ValentineJ
          Aug 22 at 11:03




          Sorry for being stupid but does "f−1" denote that it's an inverse function?
          – ValentineJ
          Aug 22 at 11:03












          yes $f^-1(x)$ is the inverse function of $f(x)$
          – Deepesh Meena
          Aug 22 at 11:03





          yes $f^-1(x)$ is the inverse function of $f(x)$
          – Deepesh Meena
          Aug 22 at 11:03













          awesome, great explanation thank you
          – ValentineJ
          Aug 22 at 11:04




          awesome, great explanation thank you
          – ValentineJ
          Aug 22 at 11:04










          up vote
          1
          down vote













          The function $f:mathbb R_geq0tomathbb R$ prescribed by $xmapstosqrt x$ has no inverse.



          This because it is not surjective.



          The function $g:mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapstosqrt x$ is bijective, hence has an inverse.



          It is the function $mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapsto x^2$.



          For finding this function see the other answers.






          share|cite|improve this answer




















          • At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
            – drhab
            Aug 23 at 17:12















          up vote
          1
          down vote













          The function $f:mathbb R_geq0tomathbb R$ prescribed by $xmapstosqrt x$ has no inverse.



          This because it is not surjective.



          The function $g:mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapstosqrt x$ is bijective, hence has an inverse.



          It is the function $mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapsto x^2$.



          For finding this function see the other answers.






          share|cite|improve this answer




















          • At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
            – drhab
            Aug 23 at 17:12













          up vote
          1
          down vote










          up vote
          1
          down vote









          The function $f:mathbb R_geq0tomathbb R$ prescribed by $xmapstosqrt x$ has no inverse.



          This because it is not surjective.



          The function $g:mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapstosqrt x$ is bijective, hence has an inverse.



          It is the function $mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapsto x^2$.



          For finding this function see the other answers.






          share|cite|improve this answer












          The function $f:mathbb R_geq0tomathbb R$ prescribed by $xmapstosqrt x$ has no inverse.



          This because it is not surjective.



          The function $g:mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapstosqrt x$ is bijective, hence has an inverse.



          It is the function $mathbb R_geq0tomathbb R_geq0$ prescribed by $xmapsto x^2$.



          For finding this function see the other answers.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 22 at 12:04









          drhab

          88.1k541120




          88.1k541120











          • At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
            – drhab
            Aug 23 at 17:12

















          • At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
            – drhab
            Aug 23 at 17:12
















          At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
          – drhab
          Aug 23 at 17:12





          At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
          – drhab
          Aug 23 at 17:12













           

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