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Find original functions from a composite function

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0

$begingroup$

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 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $fg$ is the product of $f,g$ or their composition?
  $endgroup$
  – Shubham Johri
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, there are infinitely many solutions to your problem. Consider $f(x)=2^nx+5,g(x)=x/n$.
  $endgroup$
  – Shubham Johri
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

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 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $fg$ is the product of $f,g$ or their composition?
  $endgroup$
  – Shubham Johri
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  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
  9 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, there are infinitely many solutions to your problem. Consider $f(x)=2^nx+5,g(x)=x/n$.
  $endgroup$
  – Shubham Johri
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  9 hours ago
  

  
  
  
  

 | 
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 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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 9 hours ago

Arthur

122k7122211

edited 9 hours ago

Arthur

122k7122211

asked 9 hours ago

Daniel AmaralDaniel Amaral

63

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 9 hours ago

Daniel AmaralDaniel Amaral

63

2 Answers
2

active

oldest

votes

7

$begingroup$

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 9 hours ago

Yuval GatYuval Gat

9161213

$endgroup$

add a comment | 

4

$begingroup$

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 9 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

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7

$begingroup$

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 9 hours ago

Yuval GatYuval Gat

9161213

$endgroup$

add a comment | 

7

$begingroup$

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 9 hours ago

Yuval GatYuval Gat

9161213

$endgroup$

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 9 hours ago

Yuval GatYuval Gat

9161213

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Yuval GatYuval Gat

9161213

answered 9 hours ago

Yuval GatYuval Gat

9161213

answered 9 hours ago

Yuval GatYuval Gat

9161213

4

$begingroup$

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 9 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

4

$begingroup$

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 9 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

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 9 hours ago

Cameron BuieCameron Buie

86.6k773161

$endgroup$

share|cite|improve this answer

answered 9 hours ago

Cameron BuieCameron Buie

86.6k773161

answered 9 hours ago

Cameron BuieCameron Buie

86.6k773161

answered 9 hours ago

Cameron BuieCameron Buie

86.6k773161