Mixing
Problem in calculating inverse of function
$begingroup$
$f(x)=frac{x}{1-x^2}$
I wanted to calculate its inverse.
I done following
$f(x)=y$ and using quadratic formula evaluate x in term of y.
i.e $x=frac{-1+sqrt{1-4y^2}}{2y}$ but solution is its reciprocal .I thought hard But I think I am making silly mistake. Please Help me out .
Any Help will be appreciated
algebra-precalculus functions inverse
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
$endgroup$
add a comment |
$begingroup$
$f(x)=frac{x}{1-x^2}$
I wanted to calculate its inverse.
I done following
$f(x)=y$ and using quadratic formula evaluate x in term of y.
i.e $x=frac{-1+sqrt{1-4y^2}}{2y}$ but solution is its reciprocal .I thought hard But I think I am making silly mistake. Please Help me out .
Any Help will be appreciated
algebra-precalculus functions inverse
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
$endgroup$
$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00
add a comment |
$begingroup$
$f(x)=frac{x}{1-x^2}$
I wanted to calculate its inverse.
I done following
$f(x)=y$ and using quadratic formula evaluate x in term of y.
i.e $x=frac{-1+sqrt{1-4y^2}}{2y}$ but solution is its reciprocal .I thought hard But I think I am making silly mistake. Please Help me out .
Any Help will be appreciated
algebra-precalculus functions inverse
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
$endgroup$
$f(x)=frac{x}{1-x^2}$
I wanted to calculate its inverse.
I done following
$f(x)=y$ and using quadratic formula evaluate x in term of y.
i.e $x=frac{-1+sqrt{1-4y^2}}{2y}$ but solution is its reciprocal .I thought hard But I think I am making silly mistake. Please Help me out .
Any Help will be appreciated
algebra-precalculus functions inverse
algebra-precalculus functions inverse
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
edited Jan 30 at 11:56
José Carlos Santos
172k22132239
172k22132239
asked Jan 30 at 11:45
MathLoverMathLover
56110
asked Jan 30 at 11:45
MathLoverMathLover
56110
asked Jan 30 at 11:45
MathLoverMathLover
56110
56110
$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00
add a comment |
$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00
$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00
$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, thatbegin{align}frac{-1+sqrt{1-4y^2}}{2y}&=frac{left(-1+sqrt{1-4y^2}right)left(-1-sqrt{1-4y^2}right)}{2yleft(-1-sqrt{1-4y^2}right)}\&=frac{4y^2}{2yleft(-1-sqrt{1-4y^2}right)}\&=-frac{2y}{1+sqrt{1-4y^2}}.end{align}
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
add a comment |
$begingroup$
It should be $$x=frac{y}{1-y^2}$$ or
$$x-xy^2=y$$ or
$$xy^2+y-x=0,$$ which gives
$$y=frac{-1+sqrt{1+4x^2}}{2x}$$ or $$y=frac{-1-sqrt{1+4x^2}}{2x}.$$
Now, we need to choose the right answer.
On $(-infty,-1)$ and on $(1,+infty)$ the second function is valid,
but on $(-1,1)$ we get that $g(x)=frac{-1+sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).
You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-infty,+infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :
$$x=f(y)=dfrac{-1+sqrt{1+4y^2}}{2y} tag{1}$$
(for $y neq 0$) and $f(0)=0$.
which is one you have obtained.
But if you select another branch, for example the left one, which corresponds to a function with domain $(-infty,-1)$ and range $(0,infty)$ you would have another expression than (1) for the inverse function.
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, thatbegin{align}frac{-1+sqrt{1-4y^2}}{2y}&=frac{left(-1+sqrt{1-4y^2}right)left(-1-sqrt{1-4y^2}right)}{2yleft(-1-sqrt{1-4y^2}right)}\&=frac{4y^2}{2yleft(-1-sqrt{1-4y^2}right)}\&=-frac{2y}{1+sqrt{1-4y^2}}.end{align}
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
add a comment |
$begingroup$
No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, thatbegin{align}frac{-1+sqrt{1-4y^2}}{2y}&=frac{left(-1+sqrt{1-4y^2}right)left(-1-sqrt{1-4y^2}right)}{2yleft(-1-sqrt{1-4y^2}right)}\&=frac{4y^2}{2yleft(-1-sqrt{1-4y^2}right)}\&=-frac{2y}{1+sqrt{1-4y^2}}.end{align}
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
add a comment |
$begingroup$
No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, thatbegin{align}frac{-1+sqrt{1-4y^2}}{2y}&=frac{left(-1+sqrt{1-4y^2}right)left(-1-sqrt{1-4y^2}right)}{2yleft(-1-sqrt{1-4y^2}right)}\&=frac{4y^2}{2yleft(-1-sqrt{1-4y^2}right)}\&=-frac{2y}{1+sqrt{1-4y^2}}.end{align}
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
No, you made no mistake: your solution is correct, assuming that the domain of $f$ is $(-1,1)$. Note, however, thatbegin{align}frac{-1+sqrt{1-4y^2}}{2y}&=frac{left(-1+sqrt{1-4y^2}right)left(-1-sqrt{1-4y^2}right)}{2yleft(-1-sqrt{1-4y^2}right)}\&=frac{4y^2}{2yleft(-1-sqrt{1-4y^2}right)}\&=-frac{2y}{1+sqrt{1-4y^2}}.end{align}
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
edited Jan 30 at 21:34
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 11:52
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
add a comment |
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
See my comment to the OP and my answer.
$endgroup$
– Jean Marie
Jan 30 at 19:39
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is correct, of course. I suppose that it is the restriction to $(-1,1)$ that matters here.
$endgroup$
– José Carlos Santos
Jan 30 at 21:26
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
It is an assumption that, as an educator, I think compulsory to indicate to a student.
$endgroup$
– Jean Marie
Jan 30 at 21:32
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
$begingroup$
Nice point. I've edited my first sentence.
$endgroup$
– José Carlos Santos
Jan 30 at 21:34
add a comment |
$begingroup$
It should be $$x=frac{y}{1-y^2}$$ or
$$x-xy^2=y$$ or
$$xy^2+y-x=0,$$ which gives
$$y=frac{-1+sqrt{1+4x^2}}{2x}$$ or $$y=frac{-1-sqrt{1+4x^2}}{2x}.$$
Now, we need to choose the right answer.
On $(-infty,-1)$ and on $(1,+infty)$ the second function is valid,
but on $(-1,1)$ we get that $g(x)=frac{-1+sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
It should be $$x=frac{y}{1-y^2}$$ or
$$x-xy^2=y$$ or
$$xy^2+y-x=0,$$ which gives
$$y=frac{-1+sqrt{1+4x^2}}{2x}$$ or $$y=frac{-1-sqrt{1+4x^2}}{2x}.$$
Now, we need to choose the right answer.
On $(-infty,-1)$ and on $(1,+infty)$ the second function is valid,
but on $(-1,1)$ we get that $g(x)=frac{-1+sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
It should be $$x=frac{y}{1-y^2}$$ or
$$x-xy^2=y$$ or
$$xy^2+y-x=0,$$ which gives
$$y=frac{-1+sqrt{1+4x^2}}{2x}$$ or $$y=frac{-1-sqrt{1+4x^2}}{2x}.$$
Now, we need to choose the right answer.
On $(-infty,-1)$ and on $(1,+infty)$ the second function is valid,
but on $(-1,1)$ we get that $g(x)=frac{-1+sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
It should be $$x=frac{y}{1-y^2}$$ or
$$x-xy^2=y$$ or
$$xy^2+y-x=0,$$ which gives
$$y=frac{-1+sqrt{1+4x^2}}{2x}$$ or $$y=frac{-1-sqrt{1+4x^2}}{2x}.$$
Now, we need to choose the right answer.
On $(-infty,-1)$ and on $(1,+infty)$ the second function is valid,
but on $(-1,1)$ we get that $g(x)=frac{-1+sqrt{1+4x^2}}{2x}$ is an inverse function to $f$.
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 30 at 12:13
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 11:59
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).
You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-infty,+infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :
$$x=f(y)=dfrac{-1+sqrt{1+4y^2}}{2y} tag{1}$$
(for $y neq 0$) and $f(0)=0$.
which is one you have obtained.
But if you select another branch, for example the left one, which corresponds to a function with domain $(-infty,-1)$ and range $(0,infty)$ you would have another expression than (1) for the inverse function.
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
$endgroup$
add a comment |
$begingroup$
I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).
You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-infty,+infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :
$$x=f(y)=dfrac{-1+sqrt{1+4y^2}}{2y} tag{1}$$
(for $y neq 0$) and $f(0)=0$.
which is one you have obtained.
But if you select another branch, for example the left one, which corresponds to a function with domain $(-infty,-1)$ and range $(0,infty)$ you would have another expression than (1) for the inverse function.
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
$endgroup$
add a comment |
$begingroup$
I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).
You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-infty,+infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :
$$x=f(y)=dfrac{-1+sqrt{1+4y^2}}{2y} tag{1}$$
(for $y neq 0$) and $f(0)=0$.
which is one you have obtained.
But if you select another branch, for example the left one, which corresponds to a function with domain $(-infty,-1)$ and range $(0,infty)$ you would have another expression than (1) for the inverse function.
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
$endgroup$
I am sorry but this function, as a whole, has no inverse because it is not bijective : see figure below (the curve has three branches in black).
You have, at first, to select a branch. Consider for example the central one, which corresponds to a function from $(-1,1)$ to $(-infty,+infty)$. The resulting function (graphical representation in red, symmetrical of the central branch with respect to 1st quadrant line bissector) has equation :
$$x=f(y)=dfrac{-1+sqrt{1+4y^2}}{2y} tag{1}$$
(for $y neq 0$) and $f(0)=0$.
which is one you have obtained.
But if you select another branch, for example the left one, which corresponds to a function with domain $(-infty,-1)$ and range $(0,infty)$ you would have another expression than (1) for the inverse function.
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
edited Jan 30 at 18:17
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
answered Jan 30 at 12:44
Jean MarieJean Marie
31.3k42255
31.3k42255
add a comment |
add a comment |
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$begingroup$
No, the solution of the OP is not correct : it works only if it has been specified right at the beginning that he/she restricts the domain to $(-1,1)$. Or to $(-infty,-1)$, or to $(1,+infty)$ but in this case it is not at all the same inverse function. See my answer.
$endgroup$
– Jean Marie
Jan 30 at 18:00