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5

$begingroup$

I have the following function:

$$y=fracx^2-3x^2+7$$

and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.

However, how would I find the lower bound?

real-analysis calculus

share|cite|improve this question

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

asked Sep 29 at 12:37

user709660user709660

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
  $endgroup$
  – almagest
  Sep 29 at 12:59
  

  
  
  
  

add a comment
 | 

5

$begingroup$

I have the following function:

$$y=fracx^2-3x^2+7$$

and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.

However, how would I find the lower bound?

real-analysis calculus

share|cite|improve this question

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

asked Sep 29 at 12:37

user709660user709660

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You don't need to find the best lower bound, any more than you need to find the best upper bound. $y=1-frac10x^2+7ge-frac107$
  $endgroup$
  – almagest
  Sep 29 at 12:59
  

  
  
  
  

add a comment
 | 

$begingroup$

I have the following function:

$$y=fracx^2-3x^2+7$$

and I'm trying to determine, whether the function is bounded or not. To find the upper bound, I rewrote the function as $y=fracx^2-7+10x^2+7$, and it's obvious that upper bound is 1.

However, how would I find the lower bound?

real-analysis calculus

share|cite|improve this question

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

asked Sep 29 at 12:37

user709660user709660

$endgroup$

share|cite|improve this question

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

asked Sep 29 at 12:37

user709660user709660

share|cite|improve this question

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

asked Sep 29 at 12:37

user709660user709660

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

edited Sep 29 at 13:08

cmk

7,21055 gold badges1010 silver badges3131 bronze badges

3 Answers
3

active

oldest

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3

$begingroup$

The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.

share|cite|improve this answer

edited Sep 29 at 13:00

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

We have

$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$

share|cite|improve this answer

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

$endgroup$

add a comment
 | 

1

$begingroup$

$f(x)=fracx^2-3x^2+7$

$=fracx^2+7x^2+7-frac10x^2+7$

$=1-frac10x^2+7≥1-frac107=-frac37$

$lim_x to infty f(x)=1$

So,

$1≥y≥-frac37$

So the best upper bound is $1$.

share|cite|improve this answer

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
  $endgroup$
  – N. F. Taussig
  Sep 30 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Thanks.
  $endgroup$
  – MENZIES
  Sep 30 at 8:53
  

  
  
  
  

add a comment
 | 

Your Answer

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3

$begingroup$

The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.

share|cite|improve this answer

edited Sep 29 at 13:00

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.

share|cite|improve this answer

edited Sep 29 at 13:00

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

The graph of the function is symmetric about the $y$-axis, so we only need to think about the lower bound when $xgeq 0$. And $y$ is increasing for $xgeq 0$. So the minimum occurs when $x=0$, so the greatest lower bound is $-dfrac37$.

share|cite|improve this answer

edited Sep 29 at 13:00

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 29 at 13:00

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

answered Sep 29 at 12:45

mvpqmvpq

8341111 bronze badges

3

$begingroup$

We have

$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$

share|cite|improve this answer

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

We have

$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$

share|cite|improve this answer

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

We have

$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$

share|cite|improve this answer

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

answered Sep 29 at 13:13

useruser

112k1010 gold badges4949 silver badges103103 bronze badges

1

$begingroup$

$f(x)=fracx^2-3x^2+7$

$=fracx^2+7x^2+7-frac10x^2+7$

$=1-frac10x^2+7≥1-frac107=-frac37$

$lim_x to infty f(x)=1$

So,

$1≥y≥-frac37$

So the best upper bound is $1$.

share|cite|improve this answer

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
  $endgroup$
  – N. F. Taussig
  Sep 30 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Thanks.
  $endgroup$
  – MENZIES
  Sep 30 at 8:53
  

  
  
  
  

add a comment
 | 

1

$begingroup$

$f(x)=fracx^2-3x^2+7$

$=fracx^2+7x^2+7-frac10x^2+7$

$=1-frac10x^2+7≥1-frac107=-frac37$

$lim_x to infty f(x)=1$

So,

$1≥y≥-frac37$

So the best upper bound is $1$.

share|cite|improve this answer

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is true that the least upper bound is $1$. However, $f(x) neq 1$ for any real number $x$.
  $endgroup$
  – N. F. Taussig
  Sep 30 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're right. Thanks.
  $endgroup$
  – MENZIES
  Sep 30 at 8:53
  

  
  
  
  

add a comment
 | 

$begingroup$

$f(x)=fracx^2-3x^2+7$

$=fracx^2+7x^2+7-frac10x^2+7$

$=1-frac10x^2+7≥1-frac107=-frac37$

$lim_x to infty f(x)=1$

So,

$1≥y≥-frac37$

So the best upper bound is $1$.

share|cite|improve this answer

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges

answered Sep 29 at 13:26

MENZIESMENZIES

4377 bronze badges