Find original functions from a composite function [on hold]
$begingroup$
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
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 19 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.
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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
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 19 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
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.
$endgroup$
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
functions
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
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
edited yesterday
Arthur
122k7122211
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
asked yesterday
Daniel AmaralDaniel Amaral
113
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.
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.
Daniel Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos 19 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.
put on hold as off-topic by Eevee Trainer, Leucippus, Lord Shark the Unknown, Robert Shore, José Carlos Santos 19 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
$begingroup$
$fg$ 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
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
yesterday
$begingroup$
$fg$ 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
$begingroup$
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
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
$begingroup$
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
$begingroup$
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
$begingroup$
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
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered yesterday
Yuval GatYuval Gat
9261213
$endgroup$
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 $f(x)=32x.$
answered yesterday
Cameron BuieCameron Buie
86.8k773161
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
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.
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.
answered yesterday
Yuval GatYuval Gat
9261213
$endgroup$
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.
answered yesterday
Yuval GatYuval Gat
9261213
answered yesterday
Yuval GatYuval Gat
9261213
answered yesterday
Yuval GatYuval Gat
9261213
answered yesterday
Yuval GatYuval Gat
9261213
9261213
add a comment |
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 $f(x)=32x.$
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 $f(x)=32x.$
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 $f(x)=32x.$
answered yesterday
Cameron BuieCameron Buie
86.8k773161
$endgroup$
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 $f(x)=32x.$
answered yesterday
Cameron BuieCameron Buie
86.8k773161
answered yesterday
Cameron BuieCameron Buie
86.8k773161
answered yesterday
Cameron BuieCameron Buie
86.8k773161
answered yesterday
Cameron BuieCameron Buie
86.8k773161
86.8k773161
add a comment |
add a comment |
$begingroup$
$fg$ 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