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Find original functions from a composite function [closed]

2

$begingroup$

I have the following problem that I am stuck on:

$ f(g(x))=2^x+5 $

and I must find $f(x)$ and $g(x)$

Can anyone help?

functions

share|cite|improve this question

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos Apr 8 at 9:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Robert Shore, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $fg$ is the product of $f,g$ or their composition?
  $endgroup$
  – Shubham Johri
  Apr 7 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is their composition. Also, it is 2 to the power of (x+5). Sorry I am new here I wasn't really sure how to make it like that.
  $endgroup$
  – Daniel Amaral
  Apr 7 at 18:21
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You should probably add more context, this way it is not clear for what do you want this, for example $g(x)=x$ and $f(x)=2^x+5$ would satisfy your requirements, but I doubt it is of any use.
  $endgroup$
  – Sil
  Apr 7 at 18:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, there are infinitely many solutions to your problem. Consider $f(x)=2^nx+5,g(x)=x/n$.
  $endgroup$
  – Shubham Johri
  Apr 7 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well that's really all that was in the question. Basically it said the expression for the composition of f(x) and g(x) is that, and it asked to find expressions for f(x) and g(x)
  $endgroup$
  – Daniel Amaral
  Apr 7 at 18:25
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

I have the following problem that I am stuck on:

$ f(g(x))=2^x+5 $

and I must find $f(x)$ and $g(x)$

Can anyone help?

functions

share|cite|improve this question

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos Apr 8 at 9:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Leucippus, Robert Shore, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $fg$ is the product of $f,g$ or their composition?
  $endgroup$
  – Shubham Johri
  Apr 7 at 18:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is their composition. Also, it is 2 to the power of (x+5). Sorry I am new here I wasn't really sure how to make it like that.
  $endgroup$
  – Daniel Amaral
  Apr 7 at 18:21
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You should probably add more context, this way it is not clear for what do you want this, for example $g(x)=x$ and $f(x)=2^x+5$ would satisfy your requirements, but I doubt it is of any use.
  $endgroup$
  – Sil
  Apr 7 at 18:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, there are infinitely many solutions to your problem. Consider $f(x)=2^nx+5,g(x)=x/n$.
  $endgroup$
  – Shubham Johri
  Apr 7 at 18:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well that's really all that was in the question. Basically it said the expression for the composition of f(x) and g(x) is that, and it asked to find expressions for f(x) and g(x)
  $endgroup$
  – Daniel Amaral
  Apr 7 at 18:25
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I have the following problem that I am stuck on:

$ f(g(x))=2^x+5 $

and I must find $f(x)$ and $g(x)$

Can anyone help?

functions

share|cite|improve this question

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Apr 7 at 18:22

Arthur

123k7122211

edited Apr 7 at 18:22

Arthur

123k7122211

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

New contributor

Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Apr 7 at 18:19

Daniel AmaralDaniel Amaral

163

2 Answers
2

active

oldest

votes

8

$begingroup$

Without more context, there are, of course, many solutions. For example, a rather trivial one would be $f(x)=x, g(x)=2^x+5$ (or switch the two around) or $f(x)=2^x$ and $g(x)=x+5$. More examples include $f(x)=2^x+t, g(x)=x+5-t$ for any fixed $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.

share|cite|improve this answer

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

$endgroup$

add a comment | 

5

$begingroup$

While there are infinitely-many functions $f,g$ that would do the trick, this is probably intended to give you practice choosing fairly basic (even fundamental) examples, but without just choosing one of them to be the identity function. Clearly, we'll need one of our functions to be $2^textsomething involving x$ in order for that to happen. Ideally, we'd like it to be simply $2^x.$ But which one should it be? Well, it turns out not to matter.

If we make $f(x)=2^x,$ then $fbigl(g(x)bigr)=2^g(x),$ so we'd need $g(x)=x+5.$

On the other hand, if we made $g(x)=2^x,$ then we'd have $fbigl(g(x)bigr)=fleft(2^xright),$ but what can we do to $2^x$ to turn it into $2^x+5$? Well, remember your exponent rules: $2^x+5=2^xcdot 2^5=2^xcdot 32.$ Thus, we need $f(x)=32x.$

share|cite|improve this answer

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

$endgroup$

add a comment | 

8

$begingroup$

Without more context, there are, of course, many solutions. For example, a rather trivial one would be $f(x)=x, g(x)=2^x+5$ (or switch the two around) or $f(x)=2^x$ and $g(x)=x+5$. More examples include $f(x)=2^x+t, g(x)=x+5-t$ for any fixed $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.

share|cite|improve this answer

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

$endgroup$

add a comment | 

8

$begingroup$

Without more context, there are, of course, many solutions. For example, a rather trivial one would be $f(x)=x, g(x)=2^x+5$ (or switch the two around) or $f(x)=2^x$ and $g(x)=x+5$. More examples include $f(x)=2^x+t, g(x)=x+5-t$ for any fixed $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.

share|cite|improve this answer

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

$endgroup$

add a comment | 

$begingroup$

Without more context, there are, of course, many solutions. For example, a rather trivial one would be $f(x)=x, g(x)=2^x+5$ (or switch the two around) or $f(x)=2^x$ and $g(x)=x+5$. More examples include $f(x)=2^x+t, g(x)=x+5-t$ for any fixed $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.

share|cite|improve this answer

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

$endgroup$

share|cite|improve this answer

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

answered Apr 7 at 18:27

Yuval GatYuval Gat

9641213

5

$begingroup$

While there are infinitely-many functions $f,g$ that would do the trick, this is probably intended to give you practice choosing fairly basic (even fundamental) examples, but without just choosing one of them to be the identity function. Clearly, we'll need one of our functions to be $2^textsomething involving x$ in order for that to happen. Ideally, we'd like it to be simply $2^x.$ But which one should it be? Well, it turns out not to matter.

If we make $f(x)=2^x,$ then $fbigl(g(x)bigr)=2^g(x),$ so we'd need $g(x)=x+5.$

On the other hand, if we made $g(x)=2^x,$ then we'd have $fbigl(g(x)bigr)=fleft(2^xright),$ but what can we do to $2^x$ to turn it into $2^x+5$? Well, remember your exponent rules: $2^x+5=2^xcdot 2^5=2^xcdot 32.$ Thus, we need $f(x)=32x.$

share|cite|improve this answer

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

$endgroup$

add a comment | 

5

$begingroup$

While there are infinitely-many functions $f,g$ that would do the trick, this is probably intended to give you practice choosing fairly basic (even fundamental) examples, but without just choosing one of them to be the identity function. Clearly, we'll need one of our functions to be $2^textsomething involving x$ in order for that to happen. Ideally, we'd like it to be simply $2^x.$ But which one should it be? Well, it turns out not to matter.

If we make $f(x)=2^x,$ then $fbigl(g(x)bigr)=2^g(x),$ so we'd need $g(x)=x+5.$

On the other hand, if we made $g(x)=2^x,$ then we'd have $fbigl(g(x)bigr)=fleft(2^xright),$ but what can we do to $2^x$ to turn it into $2^x+5$? Well, remember your exponent rules: $2^x+5=2^xcdot 2^5=2^xcdot 32.$ Thus, we need $f(x)=32x.$

share|cite|improve this answer

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

$endgroup$

add a comment | 

$begingroup$

While there are infinitely-many functions $f,g$ that would do the trick, this is probably intended to give you practice choosing fairly basic (even fundamental) examples, but without just choosing one of them to be the identity function. Clearly, we'll need one of our functions to be $2^textsomething involving x$ in order for that to happen. Ideally, we'd like it to be simply $2^x.$ But which one should it be? Well, it turns out not to matter.

If we make $f(x)=2^x,$ then $fbigl(g(x)bigr)=2^g(x),$ so we'd need $g(x)=x+5.$

On the other hand, if we made $g(x)=2^x,$ then we'd have $fbigl(g(x)bigr)=fleft(2^xright),$ but what can we do to $2^x$ to turn it into $2^x+5$? Well, remember your exponent rules: $2^x+5=2^xcdot 2^5=2^xcdot 32.$ Thus, we need $f(x)=32x.$

share|cite|improve this answer

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

$endgroup$

share|cite|improve this answer

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161

answered Apr 7 at 18:34

Cameron BuieCameron Buie

86.8k773161