Mixing
Darboux Theorem Proof
$begingroup$
http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29
I'm having a bit of trouble understanding the proof of Darboux's Theorem on the IVP of derivatives.
Why should there be an extremum such that $g'(x) = 0$ from the fact that $g'(a)>0$ and $g'(b)<0$ ?
calculus
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
$endgroup$
add a comment |
$begingroup$
http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29
I'm having a bit of trouble understanding the proof of Darboux's Theorem on the IVP of derivatives.
Why should there be an extremum such that $g'(x) = 0$ from the fact that $g'(a)>0$ and $g'(b)<0$ ?
calculus
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
$endgroup$
1
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09
add a comment |
$begingroup$
http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29
I'm having a bit of trouble understanding the proof of Darboux's Theorem on the IVP of derivatives.
Why should there be an extremum such that $g'(x) = 0$ from the fact that $g'(a)>0$ and $g'(b)<0$ ?
calculus
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
$endgroup$
http://en.wikipedia.org/wiki/Darboux%27s_theorem_%28analysis%29
I'm having a bit of trouble understanding the proof of Darboux's Theorem on the IVP of derivatives.
Why should there be an extremum such that $g'(x) = 0$ from the fact that $g'(a)>0$ and $g'(b)<0$ ?
calculus
calculus
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
edited Sep 7 '12 at 17:08
Siminore
30.5k33569
30.5k33569
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
asked Sep 7 '12 at 16:33
A. NapsterA. Napster
359316
359316
1
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09
add a comment |
1
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09
1
1
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose that $g$ attains a local maximum at $a$. Then $$lim_{x to a+} frac{g(x)-g(a)}{x-a} leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$lim_{x to b-} frac{g(x)-g(b)}{x-b}geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose that $g$ attains a local maximum at $a$. Then $$lim_{x to a+} frac{g(x)-g(a)}{x-a} leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$lim_{x to b-} frac{g(x)-g(b)}{x-b}geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
add a comment |
$begingroup$
Suppose that $g$ attains a local maximum at $a$. Then $$lim_{x to a+} frac{g(x)-g(a)}{x-a} leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$lim_{x to b-} frac{g(x)-g(b)}{x-b}geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
$endgroup$
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
add a comment |
$begingroup$
Suppose that $g$ attains a local maximum at $a$. Then $$lim_{x to a+} frac{g(x)-g(a)}{x-a} leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$lim_{x to b-} frac{g(x)-g(b)}{x-b}geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
$endgroup$
Suppose that $g$ attains a local maximum at $a$. Then $$lim_{x to a+} frac{g(x)-g(a)}{x-a} leq 0.$$ Analogously, if $g$ attains a local maximum at $b$, then $$lim_{x to b-} frac{g(x)-g(b)}{x-b}geq 0.$$ But both contradict $g'(a)>0$ and $g'(b)<0$. Hence the maximum, which exists since $g$ is continuous on $[a,b]$, must lie inside $[a,b]$.
Actually, you can visualize the setting: $g'(a)>0$ means that $g$ leaves $a$ in an increasing way, and then reaches $b$ in a decreasing way. Therefore $g$ must attain a maximum somewhere between $a$ and $b$.
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
answered Sep 7 '12 at 17:07
SiminoreSiminore
30.5k33569
30.5k33569
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
add a comment |
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
$begingroup$
+Upvote for the intuition in your final paragraph
$endgroup$
– Analysis
Mar 1 '14 at 12:18
add a comment |
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1
$begingroup$
What is $g$? The wikipedia page uses $f$ and $phi$...
$endgroup$
– Matthew Conroy
Sep 7 '12 at 16:35
$begingroup$
I can't understand what you mean by "...only a maximum".
$endgroup$
– Siminore
Sep 7 '12 at 17:09