How to solve function equations?

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Consider:
$$g(x)=x^1/3$$
$$f(g(x))=2x+3$$

Find $f(x)$

It should be fairly obvious that a solution is $f(x)=2x^3+3$

But is this solution unique? And what method should I use to solve such an equation if I don't "see" the solution?

functions

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edited Jul 22 at 16:34

Isham

10.6k3829

asked Jul 22 at 16:20

klutt

1033

add a comment | 

up vote
0
down vote

favorite

Consider:
$$g(x)=x^1/3$$
$$f(g(x))=2x+3$$

Find $f(x)$

It should be fairly obvious that a solution is $f(x)=2x^3+3$

But is this solution unique? And what method should I use to solve such an equation if I don't "see" the solution?

functions

share|cite|improve this question

edited Jul 22 at 16:34

Isham

10.6k3829

asked Jul 22 at 16:20

klutt

1033

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Consider:
$$g(x)=x^1/3$$
$$f(g(x))=2x+3$$

Find $f(x)$

It should be fairly obvious that a solution is $f(x)=2x^3+3$

But is this solution unique? And what method should I use to solve such an equation if I don't "see" the solution?

functions

share|cite|improve this question

edited Jul 22 at 16:34

Isham

10.6k3829

asked Jul 22 at 16:20

klutt

1033

Consider:
$$g(x)=x^1/3$$
$$f(g(x))=2x+3$$

Find $f(x)$

It should be fairly obvious that a solution is $f(x)=2x^3+3$

But is this solution unique? And what method should I use to solve such an equation if I don't "see" the solution?

functions

share|cite|improve this question

edited Jul 22 at 16:34

Isham

10.6k3829

asked Jul 22 at 16:20

klutt

1033

share|cite|improve this question

share|cite|improve this question

edited Jul 22 at 16:34

Isham

10.6k3829

edited Jul 22 at 16:34

Isham

10.6k3829

edited Jul 22 at 16:34

Isham

10.6k3829

10.6k3829

asked Jul 22 at 16:20

klutt

1033

asked Jul 22 at 16:20

klutt

1033

asked Jul 22 at 16:20

klutt

1033

1033

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
1
down vote



accepted

Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^-1 (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

add a comment | 

up vote
1
down vote

It suffices to invert $g$:

$$y=x^1/3iff x=y^3$$

and there is no "other" inverse.

Then

$$f(x)=f(g(g^-1(x)))=f(g(x^3))=2x^3+3.$$

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why this downvote ?
  – Yves Daoust
  Jul 22 at 16:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

Yes, the solution is unique.
You have two functions, $g:mathbbRrightarrow mathbbR$ and $f:mathbbRrightarrow mathbbR$.

For every $x$, $g(x)=sqrt[3]x$. You have the cube root function, and it is a bijection.

For every $x$, $f(g(x))=2x+3$. Then $f(sqrt[3]x)=2x+3$ for every $x$.
Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

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

up vote
1
down vote



accepted

Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^-1 (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

add a comment | 

up vote
1
down vote



accepted

Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^-1 (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^-1 (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

Suppose we have an equation $f(g(x)) =h(x) $ where $g$ and $h$ are known, and $g$ has an inverse. Defining $y=g(x)$, we have $f(y) =h(g^-1 (y))$, which is the solution. If $g$ does not have a unique inverse (that is, the inverse of $g(x) $ depends on x), then the solution for $f$ will not be unique.

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 16:27

Eddy

685312

answered Jul 22 at 16:27

Eddy

685312

answered Jul 22 at 16:27

Eddy

685312

685312

add a comment | 

add a comment | 

up vote
1
down vote

It suffices to invert $g$:

$$y=x^1/3iff x=y^3$$

and there is no "other" inverse.

Then

$$f(x)=f(g(g^-1(x)))=f(g(x^3))=2x^3+3.$$

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why this downvote ?
  – Yves Daoust
  Jul 22 at 16:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

It suffices to invert $g$:

$$y=x^1/3iff x=y^3$$

and there is no "other" inverse.

Then

$$f(x)=f(g(g^-1(x)))=f(g(x^3))=2x^3+3.$$

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why this downvote ?
  – Yves Daoust
  Jul 22 at 16:24
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

It suffices to invert $g$:

$$y=x^1/3iff x=y^3$$

and there is no "other" inverse.

Then

$$f(x)=f(g(g^-1(x)))=f(g(x^3))=2x^3+3.$$

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

It suffices to invert $g$:

$$y=x^1/3iff x=y^3$$

and there is no "other" inverse.

Then

$$f(x)=f(g(g^-1(x)))=f(g(x^3))=2x^3+3.$$

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 16:28

greedoid

26.1k93473

edited Jul 22 at 16:28

greedoid

26.1k93473

edited Jul 22 at 16:28

greedoid

26.1k93473

26.1k93473

answered Jul 22 at 16:23

Yves Daoust

111k665203

answered Jul 22 at 16:23

Yves Daoust

111k665203

answered Jul 22 at 16:23

Yves Daoust

111k665203

111k665203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why this downvote ?
  – Yves Daoust
  Jul 22 at 16:24
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Why this downvote ?
  – Yves Daoust
  Jul 22 at 16:24
  

  
  
  
  

Why this downvote ?
– Yves Daoust
Jul 22 at 16:24

Why this downvote ?
– Yves Daoust
Jul 22 at 16:24

add a comment | 

up vote
1
down vote

Yes, the solution is unique.
You have two functions, $g:mathbbRrightarrow mathbbR$ and $f:mathbbRrightarrow mathbbR$.

For every $x$, $g(x)=sqrt[3]x$. You have the cube root function, and it is a bijection.

For every $x$, $f(g(x))=2x+3$. Then $f(sqrt[3]x)=2x+3$ for every $x$.
Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

add a comment | 

up vote
1
down vote

Yes, the solution is unique.
You have two functions, $g:mathbbRrightarrow mathbbR$ and $f:mathbbRrightarrow mathbbR$.

For every $x$, $g(x)=sqrt[3]x$. You have the cube root function, and it is a bijection.

For every $x$, $f(g(x))=2x+3$. Then $f(sqrt[3]x)=2x+3$ for every $x$.
Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

add a comment | 

up vote
1
down vote

up vote
1
down vote

Yes, the solution is unique.
You have two functions, $g:mathbbRrightarrow mathbbR$ and $f:mathbbRrightarrow mathbbR$.

For every $x$, $g(x)=sqrt[3]x$. You have the cube root function, and it is a bijection.

For every $x$, $f(g(x))=2x+3$. Then $f(sqrt[3]x)=2x+3$ for every $x$.
Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

Yes, the solution is unique.
You have two functions, $g:mathbbRrightarrow mathbbR$ and $f:mathbbRrightarrow mathbbR$.

For every $x$, $g(x)=sqrt[3]x$. You have the cube root function, and it is a bijection.

For every $x$, $f(g(x))=2x+3$. Then $f(sqrt[3]x)=2x+3$ for every $x$.
Since cube root is a bijection, you can conclude that $f(x)=2x^3+3$.

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 16:32

Cristhian Grundmann

745

answered Jul 22 at 16:32

Cristhian Grundmann

745

answered Jul 22 at 16:32

Cristhian Grundmann

745

745

add a comment | 

add a comment | 

 

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