Investigating whether a function is boundedHow to determine whether to find an upper bound or lower boundProving Set bounded vs unboundedIs 1/(n+2) bounded or unbounded?Determine whether or not the improper integral converges $int_0^1 frace^xx^frac13 dx$Proving a set is boundedProve that the set of all almost upper bounds is nonempty, and bounded below.Finding upper bound of a linear functionDetermine whether a sequence is bounded aboveDetermine whether function is bounded$x$ bounded? How to know?
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Investigating whether a function is bounded
How to determine whether to find an upper bound or lower boundProving Set bounded vs unboundedIs 1/(n+2) bounded or unbounded?Determine whether or not the improper integral converges $int_0^1 frace^xx^frac13 dx$Proving a set is boundedProve that the set of all almost upper bounds is nonempty, and bounded below.Finding upper bound of a linear functionDetermine whether a sequence is bounded aboveDetermine whether function is bounded$ < 0$ bounded? How to know?
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margin-bottom:0;
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$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
real-analysis calculus
edited Sep 29 at 13:08
cmk
7,2105 gold badges10 silver badges31 bronze badges
7,2105 gold badges10 silver badges31 bronze badges
asked Sep 29 at 12:37
user709660user709660
$begingroup$
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
|
$begingroup$
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
$begingroup$
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
$begingroup$
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
|
3 Answers
3
active
oldest
votes
$begingroup$
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 $-dfrac37$.
$endgroup$
add a comment
|
$begingroup$
We have
$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$
$endgroup$
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 $1$.
$endgroup$
$begingroup$
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
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $-dfrac37$.
$endgroup$
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 $-dfrac37$.
$endgroup$
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 $-dfrac37$.
$endgroup$
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 $-dfrac37$.
edited Sep 29 at 13:00
answered Sep 29 at 12:45
mvpqmvpq
83411 bronze badges
83411 bronze badges
add a comment
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add a comment
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$begingroup$
We have
$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$
$endgroup$
add a comment
|
$begingroup$
We have
$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$
$endgroup$
add a comment
|
$begingroup$
We have
$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$
$endgroup$
We have
$$y=fracx^2-3x^2+7=1-frac10x^2+7ge 1-frac107$$
answered Sep 29 at 13:13
useruser
112k10 gold badges49 silver badges103 bronze badges
112k10 gold badges49 silver badges103 bronze badges
add a comment
|
add a comment
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$begingroup$
$f(x)=fracx^2-3x^2+7$
$=fracx^2+7x^2+7-frac10x^2+7$
$=1-frac10x^2+7≥1-frac107=-frac37$
$lim_x to infty f(x)=1$
So,
$1≥y≥-frac37$
So the best upper bound is $1$.
$endgroup$
$begingroup$
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 $1$.
$endgroup$
$begingroup$
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 $1$.
$endgroup$
$f(x)=fracx^2-3x^2+7$
$=fracx^2+7x^2+7-frac10x^2+7$
$=1-frac10x^2+7≥1-frac107=-frac37$
$lim_x to infty f(x)=1$
So,
$1≥y≥-frac37$
So the best upper bound is $1$.
answered Sep 29 at 13:26
MENZIESMENZIES
437 bronze badges
437 bronze badges
$begingroup$
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
|
$begingroup$
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
$begingroup$
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
$begingroup$
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
$begingroup$
You're right. Thanks.
$endgroup$
– MENZIES
Sep 30 at 8:53
add a comment
|
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$begingroup$
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