Finding extrema of function
$begingroup$
We have the function $$f_a(x)=frac{a-x}{ln (a-x)}, ain mathbb{R}$$
To get the domain of that function we have to consider the following restrictions: $$begin{cases}ln (a-x)neq 0 \ a-x>0 end{cases} Rightarrow begin{cases} a-xneq 1 \ a-x>0 end{cases} Rightarrow begin{cases} xneq a-1 \ x<a end{cases}$$ So the domain is $$D_f={xin mathbb{R}: x<a text{ and } xneq a-1}$$ Is this correct?
The function has to root since $f_a=0 Rightarrow a-x=0 Rightarrow x=anotin D_f$, right?
Next, I want to determine the extremas. The first derivative is $$f_a'=frac{-ln (a-x)+1}{left (ln (a-x)right )^2}$$ The root of the first derivative is $x=a-e$.
To check if this is a minimum or a maximum we need the second derivative, which is at $x=a-e$ positive, so we have a minimum.
Is everything correct so far?
Is this the only extremum or do we have to check also the points where the functions is not defined, i.e. $x=a$ and $x=a-1$ ?
calculus maxima-minima
edited Jan 12 at 9:57
Bernard
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
$endgroup$
add a comment |
$begingroup$
We have the function $$f_a(x)=frac{a-x}{ln (a-x)}, ain mathbb{R}$$
To get the domain of that function we have to consider the following restrictions: $$begin{cases}ln (a-x)neq 0 \ a-x>0 end{cases} Rightarrow begin{cases} a-xneq 1 \ a-x>0 end{cases} Rightarrow begin{cases} xneq a-1 \ x<a end{cases}$$ So the domain is $$D_f={xin mathbb{R}: x<a text{ and } xneq a-1}$$ Is this correct?
The function has to root since $f_a=0 Rightarrow a-x=0 Rightarrow x=anotin D_f$, right?
Next, I want to determine the extremas. The first derivative is $$f_a'=frac{-ln (a-x)+1}{left (ln (a-x)right )^2}$$ The root of the first derivative is $x=a-e$.
To check if this is a minimum or a maximum we need the second derivative, which is at $x=a-e$ positive, so we have a minimum.
Is everything correct so far?
Is this the only extremum or do we have to check also the points where the functions is not defined, i.e. $x=a$ and $x=a-1$ ?
calculus maxima-minima
edited Jan 12 at 9:57
Bernard
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
$endgroup$
add a comment |
$begingroup$
We have the function $$f_a(x)=frac{a-x}{ln (a-x)}, ain mathbb{R}$$
To get the domain of that function we have to consider the following restrictions: $$begin{cases}ln (a-x)neq 0 \ a-x>0 end{cases} Rightarrow begin{cases} a-xneq 1 \ a-x>0 end{cases} Rightarrow begin{cases} xneq a-1 \ x<a end{cases}$$ So the domain is $$D_f={xin mathbb{R}: x<a text{ and } xneq a-1}$$ Is this correct?
The function has to root since $f_a=0 Rightarrow a-x=0 Rightarrow x=anotin D_f$, right?
Next, I want to determine the extremas. The first derivative is $$f_a'=frac{-ln (a-x)+1}{left (ln (a-x)right )^2}$$ The root of the first derivative is $x=a-e$.
To check if this is a minimum or a maximum we need the second derivative, which is at $x=a-e$ positive, so we have a minimum.
Is everything correct so far?
Is this the only extremum or do we have to check also the points where the functions is not defined, i.e. $x=a$ and $x=a-1$ ?
calculus maxima-minima
edited Jan 12 at 9:57
Bernard
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
$endgroup$
We have the function $$f_a(x)=frac{a-x}{ln (a-x)}, ain mathbb{R}$$
To get the domain of that function we have to consider the following restrictions: $$begin{cases}ln (a-x)neq 0 \ a-x>0 end{cases} Rightarrow begin{cases} a-xneq 1 \ a-x>0 end{cases} Rightarrow begin{cases} xneq a-1 \ x<a end{cases}$$ So the domain is $$D_f={xin mathbb{R}: x<a text{ and } xneq a-1}$$ Is this correct?
The function has to root since $f_a=0 Rightarrow a-x=0 Rightarrow x=anotin D_f$, right?
Next, I want to determine the extremas. The first derivative is $$f_a'=frac{-ln (a-x)+1}{left (ln (a-x)right )^2}$$ The root of the first derivative is $x=a-e$.
To check if this is a minimum or a maximum we need the second derivative, which is at $x=a-e$ positive, so we have a minimum.
Is everything correct so far?
Is this the only extremum or do we have to check also the points where the functions is not defined, i.e. $x=a$ and $x=a-1$ ?
calculus maxima-minima
calculus maxima-minima
edited Jan 12 at 9:57
Bernard
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
edited Jan 12 at 9:57
Bernard
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
edited Jan 12 at 9:57
Bernard
119k740113
edited Jan 12 at 9:57
Bernard
119k740113
edited Jan 12 at 9:57
Bernard
119k740113
119k740113
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
asked Jan 12 at 9:44
Mary StarMary Star
3,02482268
3,02482268
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$,
$$lim_{xto -infty}f(x),quad lim_{xto (a-1)^-}f(x),quad lim_{xto (a-1)^+}f(x),quad lim_{xto a^-}f(x).$$
Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
$endgroup$
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
add a comment |
$begingroup$
The function tends to $pminfty$ in the points the functions is not defined therefore the point you attained is the only local minima. To check this graphically, you can refer to The graph of $Large {-xover ln (-x)}$
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
$begingroup$
Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
$endgroup$
– Mostafa Ayaz
Jan 12 at 10:50
add a comment |
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2 Answers
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2 Answers
2
active
oldest
votes
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votes
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oldest
votes
$begingroup$
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$,
$$lim_{xto -infty}f(x),quad lim_{xto (a-1)^-}f(x),quad lim_{xto (a-1)^+}f(x),quad lim_{xto a^-}f(x).$$
Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
$endgroup$
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
add a comment |
$begingroup$
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$,
$$lim_{xto -infty}f(x),quad lim_{xto (a-1)^-}f(x),quad lim_{xto (a-1)^+}f(x),quad lim_{xto a^-}f(x).$$
Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
$endgroup$
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
add a comment |
$begingroup$
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$,
$$lim_{xto -infty}f(x),quad lim_{xto (a-1)^-}f(x),quad lim_{xto (a-1)^+}f(x),quad lim_{xto a^-}f(x).$$
Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
$endgroup$
Everything is correct so far, but in order to complete your study of the global/local extrema you should also evaluate the limits at the boundary of the domain $D_f$,
$$lim_{xto -infty}f(x),quad lim_{xto (a-1)^-}f(x),quad lim_{xto (a-1)^+}f(x),quad lim_{xto a^-}f(x).$$
Are you interested also to determine the convexity/concavity of $f$? Then you need to discuss the sign of $f''$ in $D_f$.
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
edited Jan 12 at 9:59
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
answered Jan 12 at 9:53
Robert ZRobert Z
95.6k1065136
95.6k1065136
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
add a comment |
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have the limits $$lim_{xto -infty}f(x)=+infty,quad lim_{xto (a-1)^-}f(x)=-infty,quad lim_{xto (a-1)^+}f(x)=+infty,quad lim_{xto a^-}f(x)=0$$ or not? So do we havealso a minimum at $x=a$ ?
$endgroup$
– Mary Star
Jan 12 at 10:09
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
We have that $lim_{xto (a-1)^-}f(x)=+infty$ and $lim_{xto (a-1)^+}f(x)=-infty$the others are correct. No $x=a$ is not a local extrema point because it is not in $D_f$. $x=a$ is a maximum local point id we extend $f$ at $x=0$. $+infty$ and $-infty$ limits imply that there are no global minima. So you have just a local minimum at $a-e$.
$endgroup$
– Robert Z
Jan 12 at 10:15
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
Why do wee have a maximum at $x=a$ ? Could you explain to me that part further?
$endgroup$
– Mary Star
Jan 12 at 10:18
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
I meant if we extend $f$ at $a$ with the value $0$. Then $f$ is continuous in $(a-1,a]$ and there the derivative $f'$ is positive that is $f$ is increasing in $(a-1,a)$ and therefore $a$ is a local maximum. Without the exiension $a$ is not a local extrema because $anot in D_f$.
$endgroup$
– Robert Z
Jan 12 at 10:25
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
$begingroup$
@MaryStar Any further doubt? Looking at your recent questions it seems that you forgot that you have the opportunity to vote and/or accept one of the given answers.
$endgroup$
– Robert Z
Jan 12 at 11:11
add a comment |
$begingroup$
The function tends to $pminfty$ in the points the functions is not defined therefore the point you attained is the only local minima. To check this graphically, you can refer to The graph of $Large {-xover ln (-x)}$
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
$begingroup$
Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
$endgroup$
– Mostafa Ayaz
Jan 12 at 10:50
add a comment |
$begingroup$
The function tends to $pminfty$ in the points the functions is not defined therefore the point you attained is the only local minima. To check this graphically, you can refer to The graph of $Large {-xover ln (-x)}$
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
$begingroup$
Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
$endgroup$
– Mostafa Ayaz
Jan 12 at 10:50
add a comment |
$begingroup$
The function tends to $pminfty$ in the points the functions is not defined therefore the point you attained is the only local minima. To check this graphically, you can refer to The graph of $Large {-xover ln (-x)}$
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
$endgroup$
The function tends to $pminfty$ in the points the functions is not defined therefore the point you attained is the only local minima. To check this graphically, you can refer to The graph of $Large {-xover ln (-x)}$
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
answered Jan 12 at 9:59
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
$begingroup$
Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
$endgroup$
– Mostafa Ayaz
Jan 12 at 10:50
add a comment |
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
$begingroup$
Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
$endgroup$
– Mostafa Ayaz
Jan 12 at 10:50
$begingroup$
Do we not have $lim_{xto a^-}f(x)=0$ ?
$endgroup$
– Mary Star
Jan 12 at 10:06
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Do we not have $lim_{xto a^-}f(x)=0$ ?
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– Mary Star
Jan 12 at 10:06
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Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
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– Mostafa Ayaz
Jan 12 at 10:50
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Yes we have but the function is not even defined in $x=a$ but there we can define supremum instead of maxima. You also can refer to en.wikipedia.org/wiki/Maxima_and_minima
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– Mostafa Ayaz
Jan 12 at 10:50
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