Mixing
Solving $sqrt0.25xy^-0.75=y^0.25+1$ for $y$
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Is there any way of solving for $y$ for the following identity?
$$sqrt0.25xy^-0.75=y^0.25+1$$
Are there any tricks/tools to help with this problem? Alternatively, is it possible to approximate a function for $y$? I know that the implicit function theorem enables a local specification of a function for $y$ around some point, but are there any more general techniques?
algebra-precalculus
asked Jul 29 at 19:50
pafnuti
315114
add a comment |Â
up vote
0
down vote
favorite
Is there any way of solving for $y$ for the following identity?
$$sqrt0.25xy^-0.75=y^0.25+1$$
Are there any tricks/tools to help with this problem? Alternatively, is it possible to approximate a function for $y$? I know that the implicit function theorem enables a local specification of a function for $y$ around some point, but are there any more general techniques?
algebra-precalculus
asked Jul 29 at 19:50
pafnuti
315114
2
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there any way of solving for $y$ for the following identity?
$$sqrt0.25xy^-0.75=y^0.25+1$$
Are there any tricks/tools to help with this problem? Alternatively, is it possible to approximate a function for $y$? I know that the implicit function theorem enables a local specification of a function for $y$ around some point, but are there any more general techniques?
algebra-precalculus
asked Jul 29 at 19:50
pafnuti
315114
Is there any way of solving for $y$ for the following identity?
$$sqrt0.25xy^-0.75=y^0.25+1$$
Are there any tricks/tools to help with this problem? Alternatively, is it possible to approximate a function for $y$? I know that the implicit function theorem enables a local specification of a function for $y$ around some point, but are there any more general techniques?
algebra-precalculus
asked Jul 29 at 19:50
pafnuti
315114
edited Jul 29 at 20:22
asked Jul 29 at 19:50
pafnuti
315114
asked Jul 29 at 19:50
pafnuti
315114
asked Jul 29 at 19:50
pafnuti
315114
315114
2
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04
add a comment |Â
2
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04
2
2
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
HINT.-Your equation can be transformed into $$x=4y^3/4(y^1/4+1)^2$$ which can be noted as $f(y)=x$.
Define $g$ (using the same analytic expression that for $f$) by $$g(y)=4x^3/4(x^1/4+1)^2$$ and take the graphics of both functions.
These graphics, of $f$ and $g$, are such that they are symmetric respect to the first diagonal which show that $g$ is the inverse function of $f$.
Can you explain why?. Define for example $x=e^y$ and $y=e^x$ and you will see the same symmetry so you have again an example of functions such that one is the inverse of the other. Try to understand this.
answered Jul 29 at 23:38
Piquito
17.3k31234
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
HINT.-Your equation can be transformed into $$x=4y^3/4(y^1/4+1)^2$$ which can be noted as $f(y)=x$.
Define $g$ (using the same analytic expression that for $f$) by $$g(y)=4x^3/4(x^1/4+1)^2$$ and take the graphics of both functions.
These graphics, of $f$ and $g$, are such that they are symmetric respect to the first diagonal which show that $g$ is the inverse function of $f$.
Can you explain why?. Define for example $x=e^y$ and $y=e^x$ and you will see the same symmetry so you have again an example of functions such that one is the inverse of the other. Try to understand this.
answered Jul 29 at 23:38
Piquito
17.3k31234
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
add a comment |Â
up vote
1
down vote
HINT.-Your equation can be transformed into $$x=4y^3/4(y^1/4+1)^2$$ which can be noted as $f(y)=x$.
Define $g$ (using the same analytic expression that for $f$) by $$g(y)=4x^3/4(x^1/4+1)^2$$ and take the graphics of both functions.
These graphics, of $f$ and $g$, are such that they are symmetric respect to the first diagonal which show that $g$ is the inverse function of $f$.
Can you explain why?. Define for example $x=e^y$ and $y=e^x$ and you will see the same symmetry so you have again an example of functions such that one is the inverse of the other. Try to understand this.
answered Jul 29 at 23:38
Piquito
17.3k31234
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
add a comment |Â
up vote
1
down vote
up vote
1
down vote
HINT.-Your equation can be transformed into $$x=4y^3/4(y^1/4+1)^2$$ which can be noted as $f(y)=x$.
Define $g$ (using the same analytic expression that for $f$) by $$g(y)=4x^3/4(x^1/4+1)^2$$ and take the graphics of both functions.
These graphics, of $f$ and $g$, are such that they are symmetric respect to the first diagonal which show that $g$ is the inverse function of $f$.
Can you explain why?. Define for example $x=e^y$ and $y=e^x$ and you will see the same symmetry so you have again an example of functions such that one is the inverse of the other. Try to understand this.
answered Jul 29 at 23:38
Piquito
17.3k31234
HINT.-Your equation can be transformed into $$x=4y^3/4(y^1/4+1)^2$$ which can be noted as $f(y)=x$.
Define $g$ (using the same analytic expression that for $f$) by $$g(y)=4x^3/4(x^1/4+1)^2$$ and take the graphics of both functions.
These graphics, of $f$ and $g$, are such that they are symmetric respect to the first diagonal which show that $g$ is the inverse function of $f$.
Can you explain why?. Define for example $x=e^y$ and $y=e^x$ and you will see the same symmetry so you have again an example of functions such that one is the inverse of the other. Try to understand this.
answered Jul 29 at 23:38
Piquito
17.3k31234
answered Jul 29 at 23:38
Piquito
17.3k31234
answered Jul 29 at 23:38
Piquito
17.3k31234
answered Jul 29 at 23:38
Piquito
17.3k31234
17.3k31234
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
add a comment |Â
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
1
1
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
Above is $y=g(x)=4x^3/4(x^1/4+1)^2$ instead of $g(y)=4x^3/4(x^1/4+1)^2$. It was a typo.
– Piquito
Jul 30 at 2:53
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2866370%2fsolving-sqrt0-25xy-0-75-y0-251-for-y%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
The given equation is not a polynomial. Substitute the given values for $c$ and $r$ into your equation, and edit your post to include the expression you get only after substituting the given values into the equation. Note, you will be left with an equation with the variables $v$ and $b$. Do you know $v$? If not, then your goal will be to express the equation in the form $b = textsome expression with $v$ in it.$
– amWhy
Jul 29 at 19:59
@amWhy Yes that's my goal, to express b as a function of $v$.
– pafnuti
Jul 29 at 20:04