Ggthjy

Is "plugging out" electronic devices an American expression?

Finding files for which a command fails

Information to fellow intern about hiring?

Does a dangling wire really electrocute me if I'm standing in water?

Patience, young "Padovan"

Crop image to path created in TikZ?

Is there a name of the flying bionic bird?

Symmetry in quantum mechanics

Why was the "bread communication" in the arena of Catching Fire left out in the movie?

Prime joint compound before latex paint?

aging parents with no investments

Can a planet have a different gravitational pull depending on its location in orbit around its sun?

What does 'script /dev/null' do?

Are objects structures and/or vice versa?

Are cabin dividers used to "hide" the flex of the airplane?

Are white and non-white police officers equally likely to kill black suspects?

Could a US political party gain complete control over the government by removing checks & balances?

I see my dog run

How to move the player while also allowing forces to affect it

Domain expired, GoDaddy holds it and is asking more money

Filling an area between two curves

What is the meaning of "of trouble" in the following sentence?

COUNT(*) or MAX(id) - which is faster?

What is GPS' 19 year rollover and does it present a cybersecurity issue?

Find original functions from a composite function [on hold]

Composite functions and one to oneFinding range of transformation of function from range of originalHow to find the original function from a definite integral.How to find composite functions upto infinity for a given function?Behaviour of composition functions of a composite functionFunctions propertiesCan't figure out this odd functionComposite function simple questionMore on periodic functions - from 100 periodic functionsfinding an outside function given composite function

1

$begingroup$

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 yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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.

$endgroup$

put on hold as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos 20 hours ago

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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

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 yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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.

$endgroup$

put on hold as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos 20 hours ago

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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  


• 
  
  
  

  
  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
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  yesterday
  

  
  
  
  

 | 
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 yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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 yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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 yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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 yesterday

Arthur

122k7122211

edited yesterday

Arthur

122k7122211

asked yesterday

Daniel AmaralDaniel Amaral

113

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 yesterday

Daniel AmaralDaniel Amaral

113

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 $t$, $g(x)=2^{2x+10}, f(x)=sqrt{x}$, etc.

share|cite|improve this answer

answered yesterday

Yuval GatYuval Gat

9261213

$endgroup$

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 $2^{text{something involving }x}$ 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 yesterday

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 $t$, $g(x)=2^{2x+10}, f(x)=sqrt{x}$, etc.

share|cite|improve this answer

answered yesterday

Yuval GatYuval Gat

9261213

$endgroup$

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 $t$, $g(x)=2^{2x+10}, f(x)=sqrt{x}$, etc.

share|cite|improve this answer

answered yesterday

Yuval GatYuval Gat

9261213

$endgroup$

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 $t$, $g(x)=2^{2x+10}, f(x)=sqrt{x}$, etc.

share|cite|improve this answer

answered yesterday

Yuval GatYuval Gat

9261213

$endgroup$

share|cite|improve this answer

answered yesterday

Yuval GatYuval Gat

9261213

answered yesterday

Yuval GatYuval Gat

9261213

answered yesterday

Yuval GatYuval Gat

9261213

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 $2^{text{something involving }x}$ 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 yesterday

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 $2^{text{something involving }x}$ 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 yesterday

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 $2^{text{something involving }x}$ 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 yesterday

Cameron BuieCameron Buie

86.8k773161

$endgroup$

share|cite|improve this answer

answered yesterday

Cameron BuieCameron Buie

86.8k773161

answered yesterday

Cameron BuieCameron Buie

86.8k773161

answered yesterday

Cameron BuieCameron Buie

86.8k773161