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











Your Answer





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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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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
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