finding an outside function given composite function
$begingroup$
I have trouble finding the rule of the 'outside' function, given the composite function and inside function, I think you have to use some sort of substitution but otherwise I have no clue.
Does anyone know the answer to the following question:
If $p(x)= frac{4}{sqrt{3x-6}}$ and $ f(p(x)) = 3x-4$, find the rule of $f(x).$
Thanks!
functions
edited Dec 12 '18 at 12:14
amWhy
1
asked Dec 12 '18 at 11:38
apple321apple321
11
$endgroup$
add a comment |
$begingroup$
I have trouble finding the rule of the 'outside' function, given the composite function and inside function, I think you have to use some sort of substitution but otherwise I have no clue.
Does anyone know the answer to the following question:
If $p(x)= frac{4}{sqrt{3x-6}}$ and $ f(p(x)) = 3x-4$, find the rule of $f(x).$
Thanks!
functions
edited Dec 12 '18 at 12:14
amWhy
1
asked Dec 12 '18 at 11:38
apple321apple321
11
$endgroup$
add a comment |
$begingroup$
I have trouble finding the rule of the 'outside' function, given the composite function and inside function, I think you have to use some sort of substitution but otherwise I have no clue.
Does anyone know the answer to the following question:
If $p(x)= frac{4}{sqrt{3x-6}}$ and $ f(p(x)) = 3x-4$, find the rule of $f(x).$
Thanks!
functions
edited Dec 12 '18 at 12:14
amWhy
1
asked Dec 12 '18 at 11:38
apple321apple321
11
$endgroup$
I have trouble finding the rule of the 'outside' function, given the composite function and inside function, I think you have to use some sort of substitution but otherwise I have no clue.
Does anyone know the answer to the following question:
If $p(x)= frac{4}{sqrt{3x-6}}$ and $ f(p(x)) = 3x-4$, find the rule of $f(x).$
Thanks!
functions
functions
edited Dec 12 '18 at 12:14
amWhy
1
asked Dec 12 '18 at 11:38
apple321apple321
11
edited Dec 12 '18 at 12:14
amWhy
1
asked Dec 12 '18 at 11:38
apple321apple321
11
edited Dec 12 '18 at 12:14
amWhy
1
edited Dec 12 '18 at 12:14
amWhy
1
edited Dec 12 '18 at 12:14
amWhy
1
1
asked Dec 12 '18 at 11:38
apple321apple321
11
asked Dec 12 '18 at 11:38
apple321apple321
11
asked Dec 12 '18 at 11:38
apple321apple321
11
11
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Calculate the inverse function of $p(x)$, so that $p(q(x))=x$. Then replace $x$ in your problem with $q(x)$
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Things become clearer if you introduce an auxiliary variable $y=p(x)$. So we have
$y=frac{4}{sqrt{3x-6}} \ Rightarrow 3y^2(x-2) = 16 \ Rightarrow x = frac{16}{3y^2}+2 \ Rightarrow f(y) = 3x-4 = frac{16}{y^2}+2$
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Calculate the inverse function of $p(x)$, so that $p(q(x))=x$. Then replace $x$ in your problem with $q(x)$
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Calculate the inverse function of $p(x)$, so that $p(q(x))=x$. Then replace $x$ in your problem with $q(x)$
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
$endgroup$
add a comment |
$begingroup$
Calculate the inverse function of $p(x)$, so that $p(q(x))=x$. Then replace $x$ in your problem with $q(x)$
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
$endgroup$
Calculate the inverse function of $p(x)$, so that $p(q(x))=x$. Then replace $x$ in your problem with $q(x)$
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
answered Dec 12 '18 at 11:44
Empy2Empy2
33.5k12261
33.5k12261
add a comment |
add a comment |
$begingroup$
Things become clearer if you introduce an auxiliary variable $y=p(x)$. So we have
$y=frac{4}{sqrt{3x-6}} \ Rightarrow 3y^2(x-2) = 16 \ Rightarrow x = frac{16}{3y^2}+2 \ Rightarrow f(y) = 3x-4 = frac{16}{y^2}+2$
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
$endgroup$
add a comment |
$begingroup$
Things become clearer if you introduce an auxiliary variable $y=p(x)$. So we have
$y=frac{4}{sqrt{3x-6}} \ Rightarrow 3y^2(x-2) = 16 \ Rightarrow x = frac{16}{3y^2}+2 \ Rightarrow f(y) = 3x-4 = frac{16}{y^2}+2$
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
$endgroup$
add a comment |
$begingroup$
Things become clearer if you introduce an auxiliary variable $y=p(x)$. So we have
$y=frac{4}{sqrt{3x-6}} \ Rightarrow 3y^2(x-2) = 16 \ Rightarrow x = frac{16}{3y^2}+2 \ Rightarrow f(y) = 3x-4 = frac{16}{y^2}+2$
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
$endgroup$
Things become clearer if you introduce an auxiliary variable $y=p(x)$. So we have
$y=frac{4}{sqrt{3x-6}} \ Rightarrow 3y^2(x-2) = 16 \ Rightarrow x = frac{16}{3y^2}+2 \ Rightarrow f(y) = 3x-4 = frac{16}{y^2}+2$
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
answered Dec 12 '18 at 11:49
gandalf61gandalf61
8,719725
8,719725
add a comment |
add a comment |
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