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How to prove range of function is between $m$ and $M$ [closed]

1

$begingroup$

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

share|cite|improve this question

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

$endgroup$

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

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

add a comment | 

1

$begingroup$

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

share|cite|improve this question

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

$endgroup$

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

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

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

share|cite|improve this question

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

share|cite|improve this question

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

edited Mar 31 at 7:41

Minus One-Twelfth

3,458413

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

asked Mar 31 at 7:37

HellowhatsupHellowhatsup

305

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

$endgroup$

add a comment | 

1

$begingroup$

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 $y=f(x)$.

share|cite|improve this answer

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

$endgroup$

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.

share|cite|improve this answer

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

$endgroup$

share|cite|improve this answer

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

answered Mar 31 at 7:46

Simon GoodwinSimon Goodwin

1276

1

$begingroup$

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 $y=f(x)$.

share|cite|improve this answer

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

$endgroup$

add a comment | 

1

$begingroup$

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 $y=f(x)$.

share|cite|improve this answer

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

$endgroup$

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 $y=f(x)$.

share|cite|improve this answer

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

$endgroup$

share|cite|improve this answer

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748

answered Mar 31 at 7:47

Paul HurstPaul Hurst

85748