How do I determine the inverse function?
Clash Royale CLAN TAG#URR8PPP
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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
functions inverse-function
add a comment |Â
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
functions inverse-function
add a comment |Â
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
functions inverse-function
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
functions inverse-function
edited Aug 22 at 11:39
Andrés E. Caicedo
63.3k7152237
63.3k7152237
asked Aug 22 at 10:54
ValentineJ
51
51
add a comment |Â
add a comment |Â
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 $fcirc f^-1 = f^-1circ f = textIdentity map$.
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
add a comment |Â
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 $f^-1(x)=x^2$.
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
add a comment |Â
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
$$y=sqrtx$$
$$y^2=x$$
$$x=y^2$$
$$f^-1(x)=x^2$$
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
add a comment |Â
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.
At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
â drhab
Aug 23 at 17:12
add a comment |Â
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 $fcirc f^-1 = f^-1circ f = textIdentity map$.
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
add a comment |Â
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 $fcirc f^-1 = f^-1circ f = textIdentity map$.
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
add a comment |Â
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 $fcirc f^-1 = f^-1circ f = textIdentity map$.
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 $fcirc f^-1 = f^-1circ f = textIdentity map$.
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).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
add a comment |Â
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
Then $-1stackrelf^-1to1stackrelfto1neq-1$ so no identity (as it should).
â drhab
Aug 22 at 12:13
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
@drhab Ah yes, let me fix that!
â Bill Wallis
Aug 22 at 12:15
add a comment |Â
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 $f^-1(x)=x^2$.
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
add a comment |Â
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 $f^-1(x)=x^2$.
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
add a comment |Â
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 $f^-1(x)=x^2$.
We have $f(R)=R$ and $f$ is injective.
$y= sqrtx iff y^2=x$, hence $f^-1:R to R$ is given by $f^-1(x)=x^2$.
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 $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
add a comment |Â
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
What is $R$ here? If it is $mathbb R$ then it is not correct. We get $-1stackrelf^-1to1stackrelfto1neq1$.
â drhab
Aug 22 at 12:14
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
In the original post, the OP wrote $R$ instead of $mathbbR_geq 0$.
â Fred
Aug 22 at 13:05
add a comment |Â
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
$$y=sqrtx$$
$$y^2=x$$
$$x=y^2$$
$$f^-1(x)=x^2$$
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
add a comment |Â
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
$$y=sqrtx$$
$$y^2=x$$
$$x=y^2$$
$$f^-1(x)=x^2$$
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
add a comment |Â
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
$$y=sqrtx$$
$$y^2=x$$
$$x=y^2$$
$$f^-1(x)=x^2$$
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
$$y=sqrtx$$
$$y^2=x$$
$$x=y^2$$
$$f^-1(x)=x^2$$
answered Aug 22 at 11:01
Deepesh Meena
2,635719
2,635719
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
add a comment |Â
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
Sorry for being stupid but does "fâÂÂ1" denote that it's an inverse function?
â ValentineJ
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
yes $f^-1(x)$ is the inverse function of $f(x)$
â Deepesh Meena
Aug 22 at 11:03
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
awesome, great explanation thank you
â ValentineJ
Aug 22 at 11:04
add a comment |Â
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.
At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
â drhab
Aug 23 at 17:12
add a comment |Â
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.
At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
â drhab
Aug 23 at 17:12
add a comment |Â
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.
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.
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).
â drhab
Aug 23 at 17:12
add a comment |Â
At least be so kind to motivate your downvote. As long as you do not I will not take you serious (whoever you are).
â 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).
â 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).
â drhab
Aug 23 at 17:12
add a comment |Â
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