How to determine local extrema for $f(x) = xcdot sin(x) ^ {sin(x)}$
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I need to find the local extrema points of the following function:
$f(x) = xcdotsin(x) ^ {sin(x)}$
I was already able to derive to this function:
$f'(x) = x (ln(sin(x))+1)cos(x)sin(x)^{sin(x)}+sin(x)^ {sin(x)}$
calculus derivatives trigonometry maxima-minima
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
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|
show 5 more comments
$begingroup$
I need to find the local extrema points of the following function:
$f(x) = xcdotsin(x) ^ {sin(x)}$
I was already able to derive to this function:
$f'(x) = x (ln(sin(x))+1)cos(x)sin(x)^{sin(x)}+sin(x)^ {sin(x)}$
calculus derivatives trigonometry maxima-minima
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
$endgroup$
$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
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This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
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You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53
|
show 5 more comments
$begingroup$
I need to find the local extrema points of the following function:
$f(x) = xcdotsin(x) ^ {sin(x)}$
I was already able to derive to this function:
$f'(x) = x (ln(sin(x))+1)cos(x)sin(x)^{sin(x)}+sin(x)^ {sin(x)}$
calculus derivatives trigonometry maxima-minima
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
$endgroup$
I need to find the local extrema points of the following function:
$f(x) = xcdotsin(x) ^ {sin(x)}$
I was already able to derive to this function:
$f'(x) = x (ln(sin(x))+1)cos(x)sin(x)^{sin(x)}+sin(x)^ {sin(x)}$
calculus derivatives trigonometry maxima-minima
calculus derivatives trigonometry maxima-minima
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
edited Nov 23 '18 at 20:02
David G. Stork
10.2k21332
10.2k21332
asked Nov 23 '18 at 19:40
PhinsPhins
132
asked Nov 23 '18 at 19:40
PhinsPhins
132
asked Nov 23 '18 at 19:40
PhinsPhins
132
132
$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
$begingroup$
This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
$begingroup$
You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53
|
show 5 more comments
$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
$begingroup$
This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
$begingroup$
You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53
$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
$begingroup$
This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
$begingroup$
This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
$begingroup$
You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53
|
show 5 more comments
1 Answer
1
active
oldest
votes
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The domain of $f$ is where $sin x>0$ i.e. $$bigcup_{nin Bbb Z}(2npi ,2npi +pi)$$ on this domain by equaling the derivative to zero we obtain $$sin x^{sin x}=0\text{or}\ x(1+ln sin x)cos x+1=0$$where $sin x^{sin x}=0$ is always impossible and the second equation can only be solved numerically. Here is a sketch of the function
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
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add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The domain of $f$ is where $sin x>0$ i.e. $$bigcup_{nin Bbb Z}(2npi ,2npi +pi)$$ on this domain by equaling the derivative to zero we obtain $$sin x^{sin x}=0\text{or}\ x(1+ln sin x)cos x+1=0$$where $sin x^{sin x}=0$ is always impossible and the second equation can only be solved numerically. Here is a sketch of the function
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
add a comment |
$begingroup$
The domain of $f$ is where $sin x>0$ i.e. $$bigcup_{nin Bbb Z}(2npi ,2npi +pi)$$ on this domain by equaling the derivative to zero we obtain $$sin x^{sin x}=0\text{or}\ x(1+ln sin x)cos x+1=0$$where $sin x^{sin x}=0$ is always impossible and the second equation can only be solved numerically. Here is a sketch of the function
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
add a comment |
$begingroup$
The domain of $f$ is where $sin x>0$ i.e. $$bigcup_{nin Bbb Z}(2npi ,2npi +pi)$$ on this domain by equaling the derivative to zero we obtain $$sin x^{sin x}=0\text{or}\ x(1+ln sin x)cos x+1=0$$where $sin x^{sin x}=0$ is always impossible and the second equation can only be solved numerically. Here is a sketch of the function
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
$endgroup$
The domain of $f$ is where $sin x>0$ i.e. $$bigcup_{nin Bbb Z}(2npi ,2npi +pi)$$ on this domain by equaling the derivative to zero we obtain $$sin x^{sin x}=0\text{or}\ x(1+ln sin x)cos x+1=0$$where $sin x^{sin x}=0$ is always impossible and the second equation can only be solved numerically. Here is a sketch of the function
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
answered Nov 23 '18 at 20:19
Mostafa AyazMostafa Ayaz
14.7k3938
14.7k3938
add a comment |
add a comment |
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$begingroup$
find $x$ where $f'(x)$ is equal to $0$. Then assess the sign change while crossing the roots
$endgroup$
– Makina
Nov 23 '18 at 19:45
$begingroup$
This function is defined in $bigcuplimits_{ninmathbb Z}[2npi,2npi+pi]$.
$endgroup$
– Federico
Nov 23 '18 at 19:50
$begingroup$
You first have to take into considerations the extreme points of the intervals
$endgroup$
– Federico
Nov 23 '18 at 19:51
$begingroup$
Then your equation $f'=0$ inside these intervals is highly non-linear and trascendental...
$endgroup$
– Federico
Nov 23 '18 at 19:52
$begingroup$
I don't think much can be said, except numerically
$endgroup$
– Federico
Nov 23 '18 at 19:53