What is the image of the following function? [on hold]

Clash Royale CLAN TAG#URR8PPP

up vote
-1
down vote

favorite

1

So I have this.
$$g:mathbbRrightarrow mathbbR\
g(x)=2sin^2x -2sin x +3$$
What's the function's image $Im(g)$ equals to?

functions

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

put on hold as off-topic by Did, Holo, amWhy, Xander Henderson, Deepesh Meena 13 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Holo, amWhy, Xander Henderson, Deepesh Meena

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
  – Umberto P.
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
  – José Carlos Santos
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for advice
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

So I have this.
$$g:mathbbRrightarrow mathbbR\
g(x)=2sin^2x -2sin x +3$$
What's the function's image $Im(g)$ equals to?

functions

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

put on hold as off-topic by Did, Holo, amWhy, Xander Henderson, Deepesh Meena 13 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Holo, amWhy, Xander Henderson, Deepesh Meena

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
  – Umberto P.
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
  – José Carlos Santos
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for advice
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
-1
down vote

favorite

1

up vote
-1
down vote

favorite

1

1

So I have this.
$$g:mathbbRrightarrow mathbbR\
g(x)=2sin^2x -2sin x +3$$
What's the function's image $Im(g)$ equals to?

functions

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

So I have this.
$$g:mathbbRrightarrow mathbbR\
g(x)=2sin^2x -2sin x +3$$
What's the function's image $Im(g)$ equals to?

functions

functions

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

share|cite|improve this question

edited 21 hours ago

gimusi

71.6k73787

edited 21 hours ago

gimusi

71.6k73787

edited 21 hours ago

gimusi

71.6k73787

71.6k73787

asked 21 hours ago

Mihai Robert

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 21 hours ago

Mihai Robert

42

asked 21 hours ago

Mihai Robert

42

42

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Mihai Robert is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

put on hold as off-topic by Did, Holo, amWhy, Xander Henderson, Deepesh Meena 13 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Holo, amWhy, Xander Henderson, Deepesh Meena

If this question can be reworded to fit the rules in the help center, please edit the question.

put on hold as off-topic by Did, Holo, amWhy, Xander Henderson, Deepesh Meena 13 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Did, Holo, amWhy, Xander Henderson, Deepesh Meena

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
  – Umberto P.
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
  – José Carlos Santos
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for advice
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
  – Umberto P.
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
  – José Carlos Santos
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for advice
  – Mihai Robert
  21 hours ago
  

  
  
  
  

Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
– Umberto P.
21 hours ago

Since $g$ is continuous its image is an interval. Since it attains both maximum and minimum values its image is a closed interval bounded by those values. You can use elementary algebra and trig to find the max and min.
– Umberto P.
21 hours ago

1

1

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
21 hours ago

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments.
– José Carlos Santos
21 hours ago

Thank you for advice
– Mihai Robert
21 hours ago

Thank you for advice
– Mihai Robert
21 hours ago

add a comment | 

3 Answers
3

active

oldest

votes

up vote
0
down vote



accepted

HINT

We have that $g(x)$ is periodic with period $2pi$ and continuos and on $[0,2pi]$ therefore $g(x)$ has a maximum and a minimum on that interval, then consider

$$g(x)=2sin^2x -2sin x +3implies g'(x)=4sin x cos x-2 cos x=2cos x(2sin x-1)$$

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  well i've done that, but after this, everything become unclear. How do I find the max and the min ?
  – Mihai Robert
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
  – gimusi
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
  – gimusi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh, I understood. Thank you !
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
7
down vote

Hint: Write your function in the form
$$2left(sin^2(x)-sin(x)+frac14+frac32-frac14right)=2left(sin(x)-frac12right)^2+frac52$$

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

add a comment | 

up vote
0
down vote

Hint:

If you maximize and minimize the function using 1st and 2nd derivative of it, then you will find that, the maximum is 7 and the minimum is $dfrac52$. And as it is a continuous periodic function,so it's value varies from $dfrac52$ to $7$.

so the image of the function is [$dfrac52,7$].

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

HINT

We have that $g(x)$ is periodic with period $2pi$ and continuos and on $[0,2pi]$ therefore $g(x)$ has a maximum and a minimum on that interval, then consider

$$g(x)=2sin^2x -2sin x +3implies g'(x)=4sin x cos x-2 cos x=2cos x(2sin x-1)$$

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  well i've done that, but after this, everything become unclear. How do I find the max and the min ?
  – Mihai Robert
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
  – gimusi
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
  – gimusi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh, I understood. Thank you !
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

HINT

We have that $g(x)$ is periodic with period $2pi$ and continuos and on $[0,2pi]$ therefore $g(x)$ has a maximum and a minimum on that interval, then consider

$$g(x)=2sin^2x -2sin x +3implies g'(x)=4sin x cos x-2 cos x=2cos x(2sin x-1)$$

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  well i've done that, but after this, everything become unclear. How do I find the max and the min ?
  – Mihai Robert
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
  – gimusi
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
  – gimusi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh, I understood. Thank you !
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

HINT

We have that $g(x)$ is periodic with period $2pi$ and continuos and on $[0,2pi]$ therefore $g(x)$ has a maximum and a minimum on that interval, then consider

$$g(x)=2sin^2x -2sin x +3implies g'(x)=4sin x cos x-2 cos x=2cos x(2sin x-1)$$

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

HINT

We have that $g(x)$ is periodic with period $2pi$ and continuos and on $[0,2pi]$ therefore $g(x)$ has a maximum and a minimum on that interval, then consider

$$g(x)=2sin^2x -2sin x +3implies g'(x)=4sin x cos x-2 cos x=2cos x(2sin x-1)$$

share|cite|improve this answer

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

share|cite|improve this answer

share|cite|improve this answer

edited 21 hours ago

edited 21 hours ago

edited 21 hours ago

answered 21 hours ago

gimusi

71.6k73787

answered 21 hours ago

gimusi

71.6k73787

answered 21 hours ago

gimusi

71.6k73787

71.6k73787

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  well i've done that, but after this, everything become unclear. How do I find the max and the min ?
  – Mihai Robert
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
  – gimusi
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
  – gimusi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh, I understood. Thank you !
  – Mihai Robert
  21 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  well i've done that, but after this, everything become unclear. How do I find the max and the min ?
  – Mihai Robert
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
  – gimusi
  21 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
  – gimusi
  21 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Oh, I understood. Thank you !
  – Mihai Robert
  21 hours ago
  

  
  
  
  

1

1

well i've done that, but after this, everything become unclear. How do I find the max and the min ?
– Mihai Robert
21 hours ago

well i've done that, but after this, everything become unclear. How do I find the max and the min ?
– Mihai Robert
21 hours ago

1

1

@MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
– gimusi
21 hours ago

@MihaiRobert The points to consider are for $cos x=0$ $x=pi/2$ that is a local maximum and $x=3pi/2$ that is also local maximum and then for $sin x=1/2$ $x=pi/6$ that is a local minimum and $x=5pi/6$ that is also local minimum.
– gimusi
21 hours ago

1

1

@MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
– gimusi
21 hours ago

@MihaiRobert The method indicated by Dr. Sonnhard Graubner seems more effective.
– gimusi
21 hours ago

1

1

Oh, I understood. Thank you !
– Mihai Robert
21 hours ago

Oh, I understood. Thank you !
– Mihai Robert
21 hours ago

add a comment | 

up vote
7
down vote

Hint: Write your function in the form
$$2left(sin^2(x)-sin(x)+frac14+frac32-frac14right)=2left(sin(x)-frac12right)^2+frac52$$

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

add a comment | 

up vote
7
down vote

Hint: Write your function in the form
$$2left(sin^2(x)-sin(x)+frac14+frac32-frac14right)=2left(sin(x)-frac12right)^2+frac52$$

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

add a comment | 

up vote
7
down vote

up vote
7
down vote

Hint: Write your function in the form
$$2left(sin^2(x)-sin(x)+frac14+frac32-frac14right)=2left(sin(x)-frac12right)^2+frac52$$

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

Hint: Write your function in the form
$$2left(sin^2(x)-sin(x)+frac14+frac32-frac14right)=2left(sin(x)-frac12right)^2+frac52$$

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

answered 21 hours ago

Dr. Sonnhard Graubner

68.4k32760

68.4k32760

add a comment | 

add a comment | 

up vote
0
down vote

Hint:

If you maximize and minimize the function using 1st and 2nd derivative of it, then you will find that, the maximum is 7 and the minimum is $dfrac52$. And as it is a continuous periodic function,so it's value varies from $dfrac52$ to $7$.

so the image of the function is [$dfrac52,7$].

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

add a comment | 

up vote
0
down vote

Hint:

If you maximize and minimize the function using 1st and 2nd derivative of it, then you will find that, the maximum is 7 and the minimum is $dfrac52$. And as it is a continuous periodic function,so it's value varies from $dfrac52$ to $7$.

so the image of the function is [$dfrac52,7$].

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

add a comment | 

up vote
0
down vote

up vote
0
down vote

Hint:

If you maximize and minimize the function using 1st and 2nd derivative of it, then you will find that, the maximum is 7 and the minimum is $dfrac52$. And as it is a continuous periodic function,so it's value varies from $dfrac52$ to $7$.

so the image of the function is [$dfrac52,7$].

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

Hint:

If you maximize and minimize the function using 1st and 2nd derivative of it, then you will find that, the maximum is 7 and the minimum is $dfrac52$. And as it is a continuous periodic function,so it's value varies from $dfrac52$ to $7$.

so the image of the function is [$dfrac52,7$].

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

share|cite|improve this answer

share|cite|improve this answer

answered 21 hours ago

Rakibul Islam Prince

3498

answered 21 hours ago

Rakibul Islam Prince

3498

answered 21 hours ago

Rakibul Islam Prince

3498

3498

add a comment | 

add a comment |