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Find inverse of a function $t = frac1sqrt1+x^2$
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I have a formula $t = frac1sqrt1+x^2$
How is it possible to convert it into $x = +-fracsqrt1-t^2t$
I am assuming that it is an inverse function that is calculated by replacing x with t in the original equation, and then solving for x?
But I can't figure out how did it become like this. Am I just failing with the basic algebra, or is there something else to it?
inverse-function
asked Jul 25 at 21:19
Evgeniy Demidov
103
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up vote
0
down vote
favorite
I have a formula $t = frac1sqrt1+x^2$
How is it possible to convert it into $x = +-fracsqrt1-t^2t$
I am assuming that it is an inverse function that is calculated by replacing x with t in the original equation, and then solving for x?
But I can't figure out how did it become like this. Am I just failing with the basic algebra, or is there something else to it?
inverse-function
asked Jul 25 at 21:19
Evgeniy Demidov
103
1
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have a formula $t = frac1sqrt1+x^2$
How is it possible to convert it into $x = +-fracsqrt1-t^2t$
I am assuming that it is an inverse function that is calculated by replacing x with t in the original equation, and then solving for x?
But I can't figure out how did it become like this. Am I just failing with the basic algebra, or is there something else to it?
inverse-function
asked Jul 25 at 21:19
Evgeniy Demidov
103
I have a formula $t = frac1sqrt1+x^2$
How is it possible to convert it into $x = +-fracsqrt1-t^2t$
I am assuming that it is an inverse function that is calculated by replacing x with t in the original equation, and then solving for x?
But I can't figure out how did it become like this. Am I just failing with the basic algebra, or is there something else to it?
inverse-function
asked Jul 25 at 21:19
Evgeniy Demidov
103
asked Jul 25 at 21:19
Evgeniy Demidov
103
asked Jul 25 at 21:19
Evgeniy Demidov
103
asked Jul 25 at 21:19
Evgeniy Demidov
103
103
1
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36
add a comment |Â
1
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36
1
1
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36
add a comment |Â
6 Answers
6
active
oldest
votes
up vote
1
down vote
accepted
Square both sides to get
$$ t^2=frac 11+x^2$$
Reciprocate to get
$$ frac 1t^2=1+x^2$$
$$x^2=frac 1t^2-1=frac1-t^2t^2$$
Take square root and switch $x$ and $t$
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
0
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$t = frac1sqrt1+x^2$
$t^2 = frac11+x^2$
$1+x^2 = frac1t^2$
$x^2 = frac1-t^2t^2$
$x = pm sqrtfrac1-t^2t^2 = pm fracsqrt1-t^2t$ , since t > 0
answered Jul 25 at 21:42
tdluong
1315
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up vote
0
down vote
We have $0<tle 1$ and then
$$t = frac1sqrt1+x^2 iff 1+x^2=frac1t^2 iff x^2=frac1-t^2t^2$$
and then assuming $xge 0$
$$x=fracsqrt1-t^2t$$
or as an alternative, assuming $xle 0$
$$x=-fracsqrt1-t^2t$$
answered Jul 25 at 21:38
gimusi
65k73583
add a comment |Â
up vote
0
down vote
$t = frac1sqrt1+x^2$
$tsqrt1+x^2 = 1$
$sqrt1+x^2 = frac1t$
$1 + x^2 = frac1t^2$
$x^2 = frac1t^2 - 1$
$x^2 = frac1 - t^2t^2)$
$x = pmsqrtfrac1-t^2t^2$
$x = pmfracsqrt1-t^2t$
answered Jul 25 at 21:46
Phil H
1,8232311
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up vote
0
down vote
$t^2 =frac11+x^2$ $implies 1+x^2 =frac1t^2$
          $ implies x^2=frac1t^2 -1$
i.e $x^2=frac1-t^2t^2 implies $ $x = +-fracsqrt1-t^2t$
answered Jul 25 at 21:39
seifedd
424
add a comment |Â
up vote
0
down vote
Let us transform the initial equation and try to preserve the original solutions:
beginalign
t &= frac1sqrt1 + x^2 iff \
t^2 &= frac11+x^2 quad wedge quad t > 0 iff \
1 + x^2 &= frac1t^2 quad wedge quad t > 0 iff \
x^2 &= frac1t^2 - 1 quad wedge quad t > 0 iff \
x &= pmsqrtfrac1t^2 - 1 quad wedge quad t > 0 iff \
endalign
Now let us check it:
(Large version)
The green curve is the original function, with $t$ on the $y$ axis.
The blue curve is just the relation of the last line, positive root, both signs for $t$.
The orange curve is the last line, negative root, both signs for $t$.
$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$,
that indeed $t > 0$ is the correct choice for $t$.
answered Jul 25 at 21:51
mvw
30.2k22250
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Square both sides to get
$$ t^2=frac 11+x^2$$
Reciprocate to get
$$ frac 1t^2=1+x^2$$
$$x^2=frac 1t^2-1=frac1-t^2t^2$$
Take square root and switch $x$ and $t$
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
1
down vote
accepted
Square both sides to get
$$ t^2=frac 11+x^2$$
Reciprocate to get
$$ frac 1t^2=1+x^2$$
$$x^2=frac 1t^2-1=frac1-t^2t^2$$
Take square root and switch $x$ and $t$
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Square both sides to get
$$ t^2=frac 11+x^2$$
Reciprocate to get
$$ frac 1t^2=1+x^2$$
$$x^2=frac 1t^2-1=frac1-t^2t^2$$
Take square root and switch $x$ and $t$
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
Square both sides to get
$$ t^2=frac 11+x^2$$
Reciprocate to get
$$ frac 1t^2=1+x^2$$
$$x^2=frac 1t^2-1=frac1-t^2t^2$$
Take square root and switch $x$ and $t$
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 21:40
Mohammad Riazi-Kermani
27.4k41852
27.4k41852
add a comment |Â
add a comment |Â
up vote
0
down vote
$t = frac1sqrt1+x^2$
$t^2 = frac11+x^2$
$1+x^2 = frac1t^2$
$x^2 = frac1-t^2t^2$
$x = pm sqrtfrac1-t^2t^2 = pm fracsqrt1-t^2t$ , since t > 0
answered Jul 25 at 21:42
tdluong
1315
add a comment |Â
up vote
0
down vote
$t = frac1sqrt1+x^2$
$t^2 = frac11+x^2$
$1+x^2 = frac1t^2$
$x^2 = frac1-t^2t^2$
$x = pm sqrtfrac1-t^2t^2 = pm fracsqrt1-t^2t$ , since t > 0
answered Jul 25 at 21:42
tdluong
1315
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$t = frac1sqrt1+x^2$
$t^2 = frac11+x^2$
$1+x^2 = frac1t^2$
$x^2 = frac1-t^2t^2$
$x = pm sqrtfrac1-t^2t^2 = pm fracsqrt1-t^2t$ , since t > 0
answered Jul 25 at 21:42
tdluong
1315
$t = frac1sqrt1+x^2$
$t^2 = frac11+x^2$
$1+x^2 = frac1t^2$
$x^2 = frac1-t^2t^2$
$x = pm sqrtfrac1-t^2t^2 = pm fracsqrt1-t^2t$ , since t > 0
answered Jul 25 at 21:42
tdluong
1315
answered Jul 25 at 21:42
tdluong
1315
answered Jul 25 at 21:42
tdluong
1315
answered Jul 25 at 21:42
tdluong
1315
1315
add a comment |Â
add a comment |Â
up vote
0
down vote
We have $0<tle 1$ and then
$$t = frac1sqrt1+x^2 iff 1+x^2=frac1t^2 iff x^2=frac1-t^2t^2$$
and then assuming $xge 0$
$$x=fracsqrt1-t^2t$$
or as an alternative, assuming $xle 0$
$$x=-fracsqrt1-t^2t$$
answered Jul 25 at 21:38
gimusi
65k73583
add a comment |Â
up vote
0
down vote
We have $0<tle 1$ and then
$$t = frac1sqrt1+x^2 iff 1+x^2=frac1t^2 iff x^2=frac1-t^2t^2$$
and then assuming $xge 0$
$$x=fracsqrt1-t^2t$$
or as an alternative, assuming $xle 0$
$$x=-fracsqrt1-t^2t$$
answered Jul 25 at 21:38
gimusi
65k73583
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We have $0<tle 1$ and then
$$t = frac1sqrt1+x^2 iff 1+x^2=frac1t^2 iff x^2=frac1-t^2t^2$$
and then assuming $xge 0$
$$x=fracsqrt1-t^2t$$
or as an alternative, assuming $xle 0$
$$x=-fracsqrt1-t^2t$$
answered Jul 25 at 21:38
gimusi
65k73583
We have $0<tle 1$ and then
$$t = frac1sqrt1+x^2 iff 1+x^2=frac1t^2 iff x^2=frac1-t^2t^2$$
and then assuming $xge 0$
$$x=fracsqrt1-t^2t$$
or as an alternative, assuming $xle 0$
$$x=-fracsqrt1-t^2t$$
answered Jul 25 at 21:38
gimusi
65k73583
edited Jul 25 at 21:43
answered Jul 25 at 21:38
gimusi
65k73583
answered Jul 25 at 21:38
gimusi
65k73583
answered Jul 25 at 21:38
gimusi
65k73583
65k73583
add a comment |Â
add a comment |Â
up vote
0
down vote
$t = frac1sqrt1+x^2$
$tsqrt1+x^2 = 1$
$sqrt1+x^2 = frac1t$
$1 + x^2 = frac1t^2$
$x^2 = frac1t^2 - 1$
$x^2 = frac1 - t^2t^2)$
$x = pmsqrtfrac1-t^2t^2$
$x = pmfracsqrt1-t^2t$
answered Jul 25 at 21:46
Phil H
1,8232311
add a comment |Â
up vote
0
down vote
$t = frac1sqrt1+x^2$
$tsqrt1+x^2 = 1$
$sqrt1+x^2 = frac1t$
$1 + x^2 = frac1t^2$
$x^2 = frac1t^2 - 1$
$x^2 = frac1 - t^2t^2)$
$x = pmsqrtfrac1-t^2t^2$
$x = pmfracsqrt1-t^2t$
answered Jul 25 at 21:46
Phil H
1,8232311
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$t = frac1sqrt1+x^2$
$tsqrt1+x^2 = 1$
$sqrt1+x^2 = frac1t$
$1 + x^2 = frac1t^2$
$x^2 = frac1t^2 - 1$
$x^2 = frac1 - t^2t^2)$
$x = pmsqrtfrac1-t^2t^2$
$x = pmfracsqrt1-t^2t$
answered Jul 25 at 21:46
Phil H
1,8232311
$t = frac1sqrt1+x^2$
$tsqrt1+x^2 = 1$
$sqrt1+x^2 = frac1t$
$1 + x^2 = frac1t^2$
$x^2 = frac1t^2 - 1$
$x^2 = frac1 - t^2t^2)$
$x = pmsqrtfrac1-t^2t^2$
$x = pmfracsqrt1-t^2t$
answered Jul 25 at 21:46
Phil H
1,8232311
answered Jul 25 at 21:46
Phil H
1,8232311
answered Jul 25 at 21:46
Phil H
1,8232311
answered Jul 25 at 21:46
Phil H
1,8232311
1,8232311
add a comment |Â
add a comment |Â
up vote
0
down vote
$t^2 =frac11+x^2$ $implies 1+x^2 =frac1t^2$
          $ implies x^2=frac1t^2 -1$
i.e $x^2=frac1-t^2t^2 implies $ $x = +-fracsqrt1-t^2t$
answered Jul 25 at 21:39
seifedd
424
add a comment |Â
up vote
0
down vote
$t^2 =frac11+x^2$ $implies 1+x^2 =frac1t^2$
          $ implies x^2=frac1t^2 -1$
i.e $x^2=frac1-t^2t^2 implies $ $x = +-fracsqrt1-t^2t$
answered Jul 25 at 21:39
seifedd
424
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$t^2 =frac11+x^2$ $implies 1+x^2 =frac1t^2$
          $ implies x^2=frac1t^2 -1$
i.e $x^2=frac1-t^2t^2 implies $ $x = +-fracsqrt1-t^2t$
answered Jul 25 at 21:39
seifedd
424
$t^2 =frac11+x^2$ $implies 1+x^2 =frac1t^2$
          $ implies x^2=frac1t^2 -1$
i.e $x^2=frac1-t^2t^2 implies $ $x = +-fracsqrt1-t^2t$
answered Jul 25 at 21:39
seifedd
424
edited Jul 25 at 21:50
answered Jul 25 at 21:39
seifedd
424
answered Jul 25 at 21:39
seifedd
424
answered Jul 25 at 21:39
seifedd
424
424
add a comment |Â
add a comment |Â
up vote
0
down vote
Let us transform the initial equation and try to preserve the original solutions:
beginalign
t &= frac1sqrt1 + x^2 iff \
t^2 &= frac11+x^2 quad wedge quad t > 0 iff \
1 + x^2 &= frac1t^2 quad wedge quad t > 0 iff \
x^2 &= frac1t^2 - 1 quad wedge quad t > 0 iff \
x &= pmsqrtfrac1t^2 - 1 quad wedge quad t > 0 iff \
endalign
Now let us check it:
(Large version)
The green curve is the original function, with $t$ on the $y$ axis.
The blue curve is just the relation of the last line, positive root, both signs for $t$.
The orange curve is the last line, negative root, both signs for $t$.
$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$,
that indeed $t > 0$ is the correct choice for $t$.
answered Jul 25 at 21:51
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
Let us transform the initial equation and try to preserve the original solutions:
beginalign
t &= frac1sqrt1 + x^2 iff \
t^2 &= frac11+x^2 quad wedge quad t > 0 iff \
1 + x^2 &= frac1t^2 quad wedge quad t > 0 iff \
x^2 &= frac1t^2 - 1 quad wedge quad t > 0 iff \
x &= pmsqrtfrac1t^2 - 1 quad wedge quad t > 0 iff \
endalign
Now let us check it:
(Large version)
The green curve is the original function, with $t$ on the $y$ axis.
The blue curve is just the relation of the last line, positive root, both signs for $t$.
The orange curve is the last line, negative root, both signs for $t$.
$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$,
that indeed $t > 0$ is the correct choice for $t$.
answered Jul 25 at 21:51
mvw
30.2k22250
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let us transform the initial equation and try to preserve the original solutions:
beginalign
t &= frac1sqrt1 + x^2 iff \
t^2 &= frac11+x^2 quad wedge quad t > 0 iff \
1 + x^2 &= frac1t^2 quad wedge quad t > 0 iff \
x^2 &= frac1t^2 - 1 quad wedge quad t > 0 iff \
x &= pmsqrtfrac1t^2 - 1 quad wedge quad t > 0 iff \
endalign
Now let us check it:
(Large version)
The green curve is the original function, with $t$ on the $y$ axis.
The blue curve is just the relation of the last line, positive root, both signs for $t$.
The orange curve is the last line, negative root, both signs for $t$.
$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$,
that indeed $t > 0$ is the correct choice for $t$.
answered Jul 25 at 21:51
mvw
30.2k22250
Let us transform the initial equation and try to preserve the original solutions:
beginalign
t &= frac1sqrt1 + x^2 iff \
t^2 &= frac11+x^2 quad wedge quad t > 0 iff \
1 + x^2 &= frac1t^2 quad wedge quad t > 0 iff \
x^2 &= frac1t^2 - 1 quad wedge quad t > 0 iff \
x &= pmsqrtfrac1t^2 - 1 quad wedge quad t > 0 iff \
endalign
Now let us check it:
(Large version)
The green curve is the original function, with $t$ on the $y$ axis.
The blue curve is just the relation of the last line, positive root, both signs for $t$.
The orange curve is the last line, negative root, both signs for $t$.
$t$ is here on the $x$-axis, and we see, mirroring at the line $y=x$,
that indeed $t > 0$ is the correct choice for $t$.
answered Jul 25 at 21:51
mvw
30.2k22250
edited Jul 25 at 22:07
answered Jul 25 at 21:51
mvw
30.2k22250
answered Jul 25 at 21:51
mvw
30.2k22250
answered Jul 25 at 21:51
mvw
30.2k22250
30.2k22250
add a comment |Â
add a comment |Â
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1
In order to for $t = f(x)$ to have a well-defined inverse, the function $f$ needs to be an injection. Here it is not since $f(x) = f(-x)$. To fix this you could restrict to $x ge 0$ or $x le 0$, and this would tell you which sign to choose.
– User8128
Jul 25 at 21:36