How to find x0 ∈ (a, b) such that f(x0) ≤ M(b − a)/ 2 in this case? [closed]
Let $f(x)$ be a differential function on $[a, b]$ such that $f(a) = f(b) = 0$, $f(x) > 0$ where $x in (a, b)$ and $f'(x)$ is continuous on $[a, b]$.
(a) Prove that ∃M ∈ R such that |f 0 (x)| ≤ M ∀x ∈ [a, b].
(b) Find x0 ∈ (a, b) such that f(x0) ≤ M(b − a)/ 2 .
I have already solved question (a) by using the rolles-theorem, but I stuck in question b, should I find Xo by using rolles-theorem again or I have to find another way?
calculus functions derivatives rolles-theorem
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
closed as off-topic by Nosrati, clark, RRL, user10354138, Shailesh Dec 10 '18 at 10:23
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." – Nosrati, clark, RRL, user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Let $f(x)$ be a differential function on $[a, b]$ such that $f(a) = f(b) = 0$, $f(x) > 0$ where $x in (a, b)$ and $f'(x)$ is continuous on $[a, b]$.
(a) Prove that ∃M ∈ R such that |f 0 (x)| ≤ M ∀x ∈ [a, b].
(b) Find x0 ∈ (a, b) such that f(x0) ≤ M(b − a)/ 2 .
I have already solved question (a) by using the rolles-theorem, but I stuck in question b, should I find Xo by using rolles-theorem again or I have to find another way?
calculus functions derivatives rolles-theorem
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
closed as off-topic by Nosrati, clark, RRL, user10354138, Shailesh Dec 10 '18 at 10:23
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." – Nosrati, clark, RRL, user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10
add a comment |
Let $f(x)$ be a differential function on $[a, b]$ such that $f(a) = f(b) = 0$, $f(x) > 0$ where $x in (a, b)$ and $f'(x)$ is continuous on $[a, b]$.
(a) Prove that ∃M ∈ R such that |f 0 (x)| ≤ M ∀x ∈ [a, b].
(b) Find x0 ∈ (a, b) such that f(x0) ≤ M(b − a)/ 2 .
I have already solved question (a) by using the rolles-theorem, but I stuck in question b, should I find Xo by using rolles-theorem again or I have to find another way?
calculus functions derivatives rolles-theorem
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
Let $f(x)$ be a differential function on $[a, b]$ such that $f(a) = f(b) = 0$, $f(x) > 0$ where $x in (a, b)$ and $f'(x)$ is continuous on $[a, b]$.
(a) Prove that ∃M ∈ R such that |f 0 (x)| ≤ M ∀x ∈ [a, b].
(b) Find x0 ∈ (a, b) such that f(x0) ≤ M(b − a)/ 2 .
I have already solved question (a) by using the rolles-theorem, but I stuck in question b, should I find Xo by using rolles-theorem again or I have to find another way?
calculus functions derivatives rolles-theorem
calculus functions derivatives rolles-theorem
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
edited Dec 10 '18 at 4:44
Nosrati
26.5k62353
26.5k62353
asked Dec 10 '18 at 4:39
bulldog
1
asked Dec 10 '18 at 4:39
bulldog
1
asked Dec 10 '18 at 4:39
bulldog
1
1
closed as off-topic by Nosrati, clark, RRL, user10354138, Shailesh Dec 10 '18 at 10:23
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." – Nosrati, clark, RRL, user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, clark, RRL, user10354138, Shailesh Dec 10 '18 at 10:23
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." – Nosrati, clark, RRL, user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10
add a comment |
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10
add a comment |
1 Answer
1
active
oldest
votes
a) does not follow by Rolle's Theorem. It follows from the fact any continuous function on $[a,b]$ is bounded.
b) has a stronger version: for every $x_0 in [a,b]$ we have $f(x_0) leq frac {M(b-a)} 2$. To see this consider the cases $x_0 in [a,frac {a+b} 2]$ and $x_0 in [frac {a+b} 2,b]$. In the first case $f(x_0)=f(x_0)-f(a)=f'(t) (x_0-a)$ for some $t$ (by MVT). Since $x_0-aleq frac {(b-a)} 2$ we get $f(x_0) leq frac {M(b-a)} 2$. For the second case write $f(x_0)$ as $f(x_0)-f(b)$ and use a similar argument.
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
a) does not follow by Rolle's Theorem. It follows from the fact any continuous function on $[a,b]$ is bounded.
b) has a stronger version: for every $x_0 in [a,b]$ we have $f(x_0) leq frac {M(b-a)} 2$. To see this consider the cases $x_0 in [a,frac {a+b} 2]$ and $x_0 in [frac {a+b} 2,b]$. In the first case $f(x_0)=f(x_0)-f(a)=f'(t) (x_0-a)$ for some $t$ (by MVT). Since $x_0-aleq frac {(b-a)} 2$ we get $f(x_0) leq frac {M(b-a)} 2$. For the second case write $f(x_0)$ as $f(x_0)-f(b)$ and use a similar argument.
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
add a comment |
a) does not follow by Rolle's Theorem. It follows from the fact any continuous function on $[a,b]$ is bounded.
b) has a stronger version: for every $x_0 in [a,b]$ we have $f(x_0) leq frac {M(b-a)} 2$. To see this consider the cases $x_0 in [a,frac {a+b} 2]$ and $x_0 in [frac {a+b} 2,b]$. In the first case $f(x_0)=f(x_0)-f(a)=f'(t) (x_0-a)$ for some $t$ (by MVT). Since $x_0-aleq frac {(b-a)} 2$ we get $f(x_0) leq frac {M(b-a)} 2$. For the second case write $f(x_0)$ as $f(x_0)-f(b)$ and use a similar argument.
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
add a comment |
a) does not follow by Rolle's Theorem. It follows from the fact any continuous function on $[a,b]$ is bounded.
b) has a stronger version: for every $x_0 in [a,b]$ we have $f(x_0) leq frac {M(b-a)} 2$. To see this consider the cases $x_0 in [a,frac {a+b} 2]$ and $x_0 in [frac {a+b} 2,b]$. In the first case $f(x_0)=f(x_0)-f(a)=f'(t) (x_0-a)$ for some $t$ (by MVT). Since $x_0-aleq frac {(b-a)} 2$ we get $f(x_0) leq frac {M(b-a)} 2$. For the second case write $f(x_0)$ as $f(x_0)-f(b)$ and use a similar argument.
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
a) does not follow by Rolle's Theorem. It follows from the fact any continuous function on $[a,b]$ is bounded.
b) has a stronger version: for every $x_0 in [a,b]$ we have $f(x_0) leq frac {M(b-a)} 2$. To see this consider the cases $x_0 in [a,frac {a+b} 2]$ and $x_0 in [frac {a+b} 2,b]$. In the first case $f(x_0)=f(x_0)-f(a)=f'(t) (x_0-a)$ for some $t$ (by MVT). Since $x_0-aleq frac {(b-a)} 2$ we get $f(x_0) leq frac {M(b-a)} 2$. For the second case write $f(x_0)$ as $f(x_0)-f(b)$ and use a similar argument.
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
answered Dec 10 '18 at 6:09
Kavi Rama Murthy
50.4k31854
50.4k31854
add a comment |
add a comment |
Format tip: add one "!" in front of the to show the picture.
– xbh
Dec 10 '18 at 4:42
How did you prove $a)$ using Rolle's theorem?
– clark
Dec 10 '18 at 4:50
Please use MathJax in future.
– Shaun
Dec 10 '18 at 5:01
from Rolle's theorem there is a point c between a,b so that f'(c)=0 and since f(x)>0 in(a,b), we can see f'(x) is bounded in this integral.
– bulldog
Dec 10 '18 at 6:10