How to prove range of function is between $m$ and $M$ [closed] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are absolute extrema only in continuous functions?continuous function and max/min valuesExtreme values of a continuous function on a closed connected domain …Extreme values of a continuous function on a closed connected domainAbsolute maximum and minimum value of continuous function on interval $I = [a, b]$Maximum and Minimum in a range?Minimum/Maximum of $mid x mid $Closed Interval MethodShow that $f(x+t)= f(x)$Proof of Extreme Value Theorem in Stewart's Calculus book
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How to prove range of function is between $m$ and $M$ [closed]
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Are absolute extrema only in continuous functions?continuous function and max/min valuesExtreme values of a continuous function on a closed connected domain …Extreme values of a continuous function on a closed connected domainAbsolute maximum and minimum value of continuous function on interval $I = [a, b]$Maximum and Minimum in a range?Minimum/Maximum of $mid x mid $Closed Interval MethodShow that $f(x+t)= f(x)$Proof of Extreme Value Theorem in Stewart's Calculus book
$begingroup$
The question goes like this:
If $f(x)$ is a non-constant, continuous function defined on a closed interval $[a,b]$ Then by the Extreme Value Theorem, there exist an absolute minimum $m$ and an absolute maximum $M$.
Based on this, I need to show that the range of $f$, $f(x) mid a le x le b$, is the interval $[m, M]$.
Thanks in advance!
calculus
$endgroup$
closed as off-topic by Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus Apr 1 at 1:02
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus
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$begingroup$
The question goes like this:
If $f(x)$ is a non-constant, continuous function defined on a closed interval $[a,b]$ Then by the Extreme Value Theorem, there exist an absolute minimum $m$ and an absolute maximum $M$.
Based on this, I need to show that the range of $f$, $f(x) mid a le x le b$, is the interval $[m, M]$.
Thanks in advance!
calculus
$endgroup$
closed as off-topic by Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus Apr 1 at 1:02
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus
add a comment |
$begingroup$
The question goes like this:
If $f(x)$ is a non-constant, continuous function defined on a closed interval $[a,b]$ Then by the Extreme Value Theorem, there exist an absolute minimum $m$ and an absolute maximum $M$.
Based on this, I need to show that the range of $f$, $f(x) mid a le x le b$, is the interval $[m, M]$.
Thanks in advance!
calculus
$endgroup$
The question goes like this:
If $f(x)$ is a non-constant, continuous function defined on a closed interval $[a,b]$ Then by the Extreme Value Theorem, there exist an absolute minimum $m$ and an absolute maximum $M$.
Based on this, I need to show that the range of $f$, $f(x) mid a le x le b$, is the interval $[m, M]$.
Thanks in advance!
calculus
calculus
edited Mar 31 at 7:41
Minus One-Twelfth
3,458413
3,458413
asked Mar 31 at 7:37
HellowhatsupHellowhatsup
305
305
closed as off-topic by Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus Apr 1 at 1:02
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus
closed as off-topic by Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus Apr 1 at 1:02
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Thomas Shelby, egreg, José Carlos Santos, Leucippus
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider the Intermediate value theorem, let $xin [m,M]$ and then since there are $x_1,x_2$ such that $f(x_1)=m, f(x_2)=M$ we get the desired result.
$endgroup$
add a comment |
$begingroup$
First show that the range is contained in the interval $[m,M]$. Then use the Intermediate Value Theorem to show that if $y in [m,M]$ then there exists an $x in [a,b]$ with $y=f(x)$.
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the Intermediate value theorem, let $xin [m,M]$ and then since there are $x_1,x_2$ such that $f(x_1)=m, f(x_2)=M$ we get the desired result.
$endgroup$
add a comment |
$begingroup$
Consider the Intermediate value theorem, let $xin [m,M]$ and then since there are $x_1,x_2$ such that $f(x_1)=m, f(x_2)=M$ we get the desired result.
$endgroup$
add a comment |
$begingroup$
Consider the Intermediate value theorem, let $xin [m,M]$ and then since there are $x_1,x_2$ such that $f(x_1)=m, f(x_2)=M$ we get the desired result.
$endgroup$
Consider the Intermediate value theorem, let $xin [m,M]$ and then since there are $x_1,x_2$ such that $f(x_1)=m, f(x_2)=M$ we get the desired result.
answered Mar 31 at 7:46
Simon GoodwinSimon Goodwin
1276
1276
add a comment |
add a comment |
$begingroup$
First show that the range is contained in the interval $[m,M]$. Then use the Intermediate Value Theorem to show that if $y in [m,M]$ then there exists an $x in [a,b]$ with $y=f(x)$.
$endgroup$
add a comment |
$begingroup$
First show that the range is contained in the interval $[m,M]$. Then use the Intermediate Value Theorem to show that if $y in [m,M]$ then there exists an $x in [a,b]$ with $y=f(x)$.
$endgroup$
add a comment |
$begingroup$
First show that the range is contained in the interval $[m,M]$. Then use the Intermediate Value Theorem to show that if $y in [m,M]$ then there exists an $x in [a,b]$ with $y=f(x)$.
$endgroup$
First show that the range is contained in the interval $[m,M]$. Then use the Intermediate Value Theorem to show that if $y in [m,M]$ then there exists an $x in [a,b]$ with $y=f(x)$.
answered Mar 31 at 7:47
Paul HurstPaul Hurst
85748
85748
add a comment |
add a comment |