Find the value of inverse function with $ 2 pi $ argument
1
$begingroup$
If $ f(x) = (2x-3 pi)^5 + (4/3)x + cos(x) $ and g is the inverse function of f, then what is the value of $ g'(2 pi) $
Mt try :
functions
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
$endgroup$
add a comment |
1
$begingroup$
If $ f(x) = (2x-3 pi)^5 + (4/3)x + cos(x) $ and g is the inverse function of f, then what is the value of $ g'(2 pi) $
Mt try :
functions
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
$endgroup$
3
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
1
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03
add a comment |
1
1
1
$begingroup$
If $ f(x) = (2x-3 pi)^5 + (4/3)x + cos(x) $ and g is the inverse function of f, then what is the value of $ g'(2 pi) $
Mt try :
functions
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
$endgroup$
If $ f(x) = (2x-3 pi)^5 + (4/3)x + cos(x) $ and g is the inverse function of f, then what is the value of $ g'(2 pi) $
Mt try :
functions
functions
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
edited Dec 11 '18 at 16:58
Warren Hill
2,7091022
2,7091022
asked Dec 11 '18 at 15:57
faceface
164
asked Dec 11 '18 at 15:57
faceface
164
asked Dec 11 '18 at 15:57
faceface
164
164
3
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
1
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03
add a comment |
3
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
1
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03
3
3
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
1
1
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03
add a comment |
1 Answer
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2
$begingroup$
We have $f(frac{3}{2} pi) = 2pi$, so $g(2pi) = frac{3}{2} pi$. An ugly exercise (in my opinion) because one have to guess this value.
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
$endgroup$
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
add a comment |
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1 Answer
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1 Answer
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active
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2
$begingroup$
We have $f(frac{3}{2} pi) = 2pi$, so $g(2pi) = frac{3}{2} pi$. An ugly exercise (in my opinion) because one have to guess this value.
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
$endgroup$
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
add a comment |
2
$begingroup$
We have $f(frac{3}{2} pi) = 2pi$, so $g(2pi) = frac{3}{2} pi$. An ugly exercise (in my opinion) because one have to guess this value.
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
$endgroup$
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
add a comment |
2
2
2
$begingroup$
We have $f(frac{3}{2} pi) = 2pi$, so $g(2pi) = frac{3}{2} pi$. An ugly exercise (in my opinion) because one have to guess this value.
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
$endgroup$
We have $f(frac{3}{2} pi) = 2pi$, so $g(2pi) = frac{3}{2} pi$. An ugly exercise (in my opinion) because one have to guess this value.
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
answered Dec 11 '18 at 16:20
jjagmathjjagmath
3387
3387
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
add a comment |
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
$begingroup$
Clever - I hadn't thought of this. You could probably find the value with the Lagrange Inversion Theorem if it wasn't possible by inspection.
$endgroup$
– Jam
Dec 11 '18 at 17:22
add a comment |
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3
$begingroup$
Please use Mathjax to format your equations.
$endgroup$
– Jam
Dec 11 '18 at 16:00
1
$begingroup$
You've missed out the power on $(2y-3pi)$ after implicitly differentiating.
$endgroup$
– Jam
Dec 11 '18 at 16:02
$begingroup$
Other than that, it seems fine to me :). I'm not sure you'd have a closed form for $g(2pi)$ though.
$endgroup$
– Jam
Dec 11 '18 at 16:03