Finding range of $a$












2












$begingroup$



If $$f(x) =
begin{cases}
|x-2|+a^2-9a-9, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$ has local minima at $x=2$, then range of $a$ is… ?




My failed Attempt:



Wrote
$$f(x) =
begin{cases}
-x+a^2-9a-7, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$



The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was



Then My next step would be differentiating



$$f(x) =
begin{cases}
-1, &text{if }x<2\
2, &text{if } xgeqslant2
end{cases}$$



That means the function is decreasing before $x=2$ and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    A non-continuous function can also have (local) extrema.
    $endgroup$
    – Karlo
    Sep 24 '18 at 10:34










  • $begingroup$
    I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:36






  • 2




    $begingroup$
    Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
    $endgroup$
    – Florian
    Sep 24 '18 at 10:38










  • $begingroup$
    I miswrote that when double checked with my textbook. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:41
















2












$begingroup$



If $$f(x) =
begin{cases}
|x-2|+a^2-9a-9, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$ has local minima at $x=2$, then range of $a$ is… ?




My failed Attempt:



Wrote
$$f(x) =
begin{cases}
-x+a^2-9a-7, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$



The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was



Then My next step would be differentiating



$$f(x) =
begin{cases}
-1, &text{if }x<2\
2, &text{if } xgeqslant2
end{cases}$$



That means the function is decreasing before $x=2$ and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    A non-continuous function can also have (local) extrema.
    $endgroup$
    – Karlo
    Sep 24 '18 at 10:34










  • $begingroup$
    I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:36






  • 2




    $begingroup$
    Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
    $endgroup$
    – Florian
    Sep 24 '18 at 10:38










  • $begingroup$
    I miswrote that when double checked with my textbook. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:41














2












2








2


1



$begingroup$



If $$f(x) =
begin{cases}
|x-2|+a^2-9a-9, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$ has local minima at $x=2$, then range of $a$ is… ?




My failed Attempt:



Wrote
$$f(x) =
begin{cases}
-x+a^2-9a-7, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$



The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was



Then My next step would be differentiating



$$f(x) =
begin{cases}
-1, &text{if }x<2\
2, &text{if } xgeqslant2
end{cases}$$



That means the function is decreasing before $x=2$ and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed






calculus







share|cite|improve this question











$endgroup$





If $$f(x) =
begin{cases}
|x-2|+a^2-9a-9, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$ has local minima at $x=2$, then range of $a$ is… ?




My failed Attempt:



Wrote
$$f(x) =
begin{cases}
-x+a^2-9a-7, &text{if }x<2\
2x-3, &text{if } xgeqslant2
end{cases}$$



The problem here is that I don't know whether the function is continious is required or not for finding maxima and minima. Not that I could check it if it was



Then My next step would be differentiating



$$f(x) =
begin{cases}
-1, &text{if }x<2\
2, &text{if } xgeqslant2
end{cases}$$



That means the function is decreasing before $x=2$ and then increasing. This is the best I could think for this question but it's not providing me any hint on how to proceed






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 27 at 15:02







Zerix

















asked Sep 24 '18 at 10:26









ZerixZerix

1378




1378












  • $begingroup$
    A non-continuous function can also have (local) extrema.
    $endgroup$
    – Karlo
    Sep 24 '18 at 10:34










  • $begingroup$
    I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:36






  • 2




    $begingroup$
    Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
    $endgroup$
    – Florian
    Sep 24 '18 at 10:38










  • $begingroup$
    I miswrote that when double checked with my textbook. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:41


















  • $begingroup$
    A non-continuous function can also have (local) extrema.
    $endgroup$
    – Karlo
    Sep 24 '18 at 10:34










  • $begingroup$
    I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:36






  • 2




    $begingroup$
    Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
    $endgroup$
    – Florian
    Sep 24 '18 at 10:38










  • $begingroup$
    I miswrote that when double checked with my textbook. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 10:41
















$begingroup$
A non-continuous function can also have (local) extrema.
$endgroup$
– Karlo
Sep 24 '18 at 10:34




$begingroup$
A non-continuous function can also have (local) extrema.
$endgroup$
– Karlo
Sep 24 '18 at 10:34












$begingroup$
I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
$endgroup$
– Zerix
Sep 24 '18 at 10:36




$begingroup$
I always get confused at non continious functions. I haven't seen examples of them so that's why I am always little confused when dealing with them
$endgroup$
– Zerix
Sep 24 '18 at 10:36




2




2




$begingroup$
Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
$endgroup$
– Florian
Sep 24 '18 at 10:38




$begingroup$
Should it be $|x-2|$ in the task? Otherwise I do not see how you get to $-x$ in the next step.
$endgroup$
– Florian
Sep 24 '18 at 10:38












$begingroup$
I miswrote that when double checked with my textbook. Sorry for the inconvenience
$endgroup$
– Zerix
Sep 24 '18 at 10:41




$begingroup$
I miswrote that when double checked with my textbook. Sorry for the inconvenience
$endgroup$
– Zerix
Sep 24 '18 at 10:41










2 Answers
2






active

oldest

votes


















7












$begingroup$

A non-continuous function can also have the minimum at $x=2$



Here you want a local minimum at $x=2$ thus
$$
|2-2|+a^2-9a-9ge 2cdot2-3
$$



$$
a^2-9a-9ge 1
$$
$$(a+1)(a-10)ge 0$$



You get $$ain(-infty,-1]cup[10,infty)$$



You can look here on desmos for the simulation






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 14:11



















2












$begingroup$

There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?



So, put right hand limit and you get 4-3=1.

Put x=2 for x<2 to get $a^2$-9a-9.



Since the function is continuous, both limits will be same, i.e. 1

So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$
$$(a-10)(a+1)=0 $$
$$a=10$$ or $$a=-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
    $endgroup$
    – Zerix
    Sep 24 '18 at 11:08












  • $begingroup$
    a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
    $endgroup$
    – idea
    Sep 24 '18 at 11:27












  • $begingroup$
    you solution is partially correct look at my solution hope you understand
    $endgroup$
    – Deepesh Meena
    Sep 24 '18 at 12:58











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

A non-continuous function can also have the minimum at $x=2$



Here you want a local minimum at $x=2$ thus
$$
|2-2|+a^2-9a-9ge 2cdot2-3
$$



$$
a^2-9a-9ge 1
$$
$$(a+1)(a-10)ge 0$$



You get $$ain(-infty,-1]cup[10,infty)$$



You can look here on desmos for the simulation






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 14:11
















7












$begingroup$

A non-continuous function can also have the minimum at $x=2$



Here you want a local minimum at $x=2$ thus
$$
|2-2|+a^2-9a-9ge 2cdot2-3
$$



$$
a^2-9a-9ge 1
$$
$$(a+1)(a-10)ge 0$$



You get $$ain(-infty,-1]cup[10,infty)$$



You can look here on desmos for the simulation






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 14:11














7












7








7





$begingroup$

A non-continuous function can also have the minimum at $x=2$



Here you want a local minimum at $x=2$ thus
$$
|2-2|+a^2-9a-9ge 2cdot2-3
$$



$$
a^2-9a-9ge 1
$$
$$(a+1)(a-10)ge 0$$



You get $$ain(-infty,-1]cup[10,infty)$$



You can look here on desmos for the simulation






share|cite|improve this answer











$endgroup$



A non-continuous function can also have the minimum at $x=2$



Here you want a local minimum at $x=2$ thus
$$
|2-2|+a^2-9a-9ge 2cdot2-3
$$



$$
a^2-9a-9ge 1
$$
$$(a+1)(a-10)ge 0$$



You get $$ain(-infty,-1]cup[10,infty)$$



You can look here on desmos for the simulation







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 24 '18 at 13:25

























answered Sep 24 '18 at 12:46









Deepesh MeenaDeepesh Meena

4,30421025




4,30421025












  • $begingroup$
    Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 14:11


















  • $begingroup$
    Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
    $endgroup$
    – Zerix
    Sep 24 '18 at 14:11
















$begingroup$
Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
$endgroup$
– Zerix
Sep 24 '18 at 14:11




$begingroup$
Can you tell me what did you do..... I am unable to. Understand it. Sorry for the inconvenience
$endgroup$
– Zerix
Sep 24 '18 at 14:11











2












$begingroup$

There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?



So, put right hand limit and you get 4-3=1.

Put x=2 for x<2 to get $a^2$-9a-9.



Since the function is continuous, both limits will be same, i.e. 1

So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$
$$(a-10)(a+1)=0 $$
$$a=10$$ or $$a=-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
    $endgroup$
    – Zerix
    Sep 24 '18 at 11:08












  • $begingroup$
    a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
    $endgroup$
    – idea
    Sep 24 '18 at 11:27












  • $begingroup$
    you solution is partially correct look at my solution hope you understand
    $endgroup$
    – Deepesh Meena
    Sep 24 '18 at 12:58
















2












$begingroup$

There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?



So, put right hand limit and you get 4-3=1.

Put x=2 for x<2 to get $a^2$-9a-9.



Since the function is continuous, both limits will be same, i.e. 1

So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$
$$(a-10)(a+1)=0 $$
$$a=10$$ or $$a=-1$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
    $endgroup$
    – Zerix
    Sep 24 '18 at 11:08












  • $begingroup$
    a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
    $endgroup$
    – idea
    Sep 24 '18 at 11:27












  • $begingroup$
    you solution is partially correct look at my solution hope you understand
    $endgroup$
    – Deepesh Meena
    Sep 24 '18 at 12:58














2












2








2





$begingroup$

There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?



So, put right hand limit and you get 4-3=1.

Put x=2 for x<2 to get $a^2$-9a-9.



Since the function is continuous, both limits will be same, i.e. 1

So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$
$$(a-10)(a+1)=0 $$
$$a=10$$ or $$a=-1$$






share|cite|improve this answer









$endgroup$



There's no other way of doing this question except assuming it is continuous, otherwise why would they have given you the function broken like this at x=2?



So, put right hand limit and you get 4-3=1.

Put x=2 for x<2 to get $a^2$-9a-9.



Since the function is continuous, both limits will be same, i.e. 1

So, $$a^2-9a-9=1$$ $$ a^2-9a-10=0 $$
$$(a-10)(a+1)=0 $$
$$a=10$$ or $$a=-1$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 24 '18 at 11:01









ideaidea

2,16841125




2,16841125












  • $begingroup$
    If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
    $endgroup$
    – Zerix
    Sep 24 '18 at 11:08












  • $begingroup$
    a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
    $endgroup$
    – idea
    Sep 24 '18 at 11:27












  • $begingroup$
    you solution is partially correct look at my solution hope you understand
    $endgroup$
    – Deepesh Meena
    Sep 24 '18 at 12:58


















  • $begingroup$
    If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
    $endgroup$
    – Zerix
    Sep 24 '18 at 11:08












  • $begingroup$
    a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
    $endgroup$
    – idea
    Sep 24 '18 at 11:27












  • $begingroup$
    you solution is partially correct look at my solution hope you understand
    $endgroup$
    – Deepesh Meena
    Sep 24 '18 at 12:58
















$begingroup$
If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
$endgroup$
– Zerix
Sep 24 '18 at 11:08






$begingroup$
If I substitute a=-2 in the question then I get -x+15.This gives decreasing till 2 then increasing making x=2 minima I guess so I don't think x=-1 is the only solution. And the question also asks about range
$endgroup$
– Zerix
Sep 24 '18 at 11:08














$begingroup$
a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
$endgroup$
– idea
Sep 24 '18 at 11:27






$begingroup$
a=10 and a=-1 are the 2 solutions.There are discrete values and not the range for a. And, given function needs to be decreasing and then increasing if it has to be continuous.
$endgroup$
– idea
Sep 24 '18 at 11:27














$begingroup$
you solution is partially correct look at my solution hope you understand
$endgroup$
– Deepesh Meena
Sep 24 '18 at 12:58




$begingroup$
you solution is partially correct look at my solution hope you understand
$endgroup$
– Deepesh Meena
Sep 24 '18 at 12:58


















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Le Mesnil-Réaume

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Le Mesnil-Réaume Region Normandie Département Seine-Maritime Arrondissement Dieppe Kanton Eu Gemeindeverband Villes Sœurs Koordinaten 49° 58′  N , 1° 27′  O 49.9725 1.4463888888889 Koordinaten: 49° 58′  N , 1° 27′  O Höhe 92–156 m Fläche 5,55 km 2 Einwohner 793 (1. Januar 2016) Bevölkerungsdichte 143 Einw./km 2 Postleitzahl 76260 INSEE-Code 76435 Website http://www.mesnil-reaume.net/ Rathaus (Mairie) und Schule von Le Mesnil-Réaume Le Mesnil-Réaume ist eine französische Gemeinde mit 793 Einwohnern (Stand: 1. Januar 2016) im Département Seine-Maritime in der Region Normandie (vor 2016 Haute-Normandie). Sie gehört zum Arrondissement Dieppe und zum Kanton Eu. Die Einwohner werden Mesnillais genannt. Inhaltsverzeichnis 1 Geographie 2 Bevölkerungsentwicklung 3 Sehenswürdigkeiten 4 Weblinks Geographie | Le Mesnil-Réaume liegt etwa 43 Kilomet
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