Find original functions from a composite functionComposite 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
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Find original functions from a composite function
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
$begingroup$
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
New contributor
$endgroup$
|
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
New contributor
$endgroup$
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 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
7 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
7 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
7 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
7 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
New contributor
$endgroup$
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
New contributor
New contributor
edited 7 hours ago
Arthur
122k7122211
122k7122211
New contributor
asked 7 hours ago
Daniel AmaralDaniel Amaral
63
63
New contributor
New contributor
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 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
7 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
7 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
7 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
7 hours ago
|
show 1 more comment
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 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
7 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
7 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
7 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
7 hours ago
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 hours ago
$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 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
7 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
7 hours ago
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
7 hours ago
$begingroup$
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
7 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
7 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
7 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
7 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
7 hours ago
|
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 $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.
$endgroup$
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 $f(x)=32x.$
$endgroup$
add a comment |
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2 Answers
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$begingroup$
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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 $t$, $g(x)=2^2x+10, f(x)=sqrtx$, etc.
answered 7 hours ago
Yuval GatYuval Gat
9161213
9161213
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 $2^textsomething involving x$ 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 $f(x)=32x.$
$endgroup$
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 $f(x)=32x.$
$endgroup$
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 $f(x)=32x.$
$endgroup$
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 $2^textsomething involving x$ 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 $f(x)=32x.$
answered 7 hours ago
Cameron BuieCameron Buie
86.6k773161
86.6k773161
add a comment |
add a comment |
Daniel Amaral is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Amaral is a new contributor. Be nice, and check out our Code of Conduct.
Daniel Amaral is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
$fg$ is the product of $f,g$ or their composition?
$endgroup$
– Shubham Johri
7 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
7 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
7 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
7 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
7 hours ago