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Exercise about derivability and continuity of a function
Clash Royale CLAN TAG#URR8PPP
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Given $a in mathbbR$, $f:[a,infty) to mathbbR$ continous and $lim_x to inftyf(x)=f(a)$. Knowing $f$ is derivable onto $(a,infty)$ prove that exists $x_0 > a$ so that $f'(x_0)=0$
If $f$ is a constant function the thesis is obvious. But in the other case I have no idea how to proceed.
real-analysis analysis
edited Jul 17 at 13:45
Robert Z
84.2k955123
asked Jul 17 at 13:12
user515933
846
add a comment |Â
up vote
1
down vote
favorite
Given $a in mathbbR$, $f:[a,infty) to mathbbR$ continous and $lim_x to inftyf(x)=f(a)$. Knowing $f$ is derivable onto $(a,infty)$ prove that exists $x_0 > a$ so that $f'(x_0)=0$
If $f$ is a constant function the thesis is obvious. But in the other case I have no idea how to proceed.
real-analysis analysis
edited Jul 17 at 13:45
Robert Z
84.2k955123
asked Jul 17 at 13:12
user515933
846
The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $a in mathbbR$, $f:[a,infty) to mathbbR$ continous and $lim_x to inftyf(x)=f(a)$. Knowing $f$ is derivable onto $(a,infty)$ prove that exists $x_0 > a$ so that $f'(x_0)=0$
If $f$ is a constant function the thesis is obvious. But in the other case I have no idea how to proceed.
real-analysis analysis
edited Jul 17 at 13:45
Robert Z
84.2k955123
asked Jul 17 at 13:12
user515933
846
Given $a in mathbbR$, $f:[a,infty) to mathbbR$ continous and $lim_x to inftyf(x)=f(a)$. Knowing $f$ is derivable onto $(a,infty)$ prove that exists $x_0 > a$ so that $f'(x_0)=0$
If $f$ is a constant function the thesis is obvious. But in the other case I have no idea how to proceed.
real-analysis analysis
edited Jul 17 at 13:45
Robert Z
84.2k955123
asked Jul 17 at 13:12
user515933
846
edited Jul 17 at 13:45
Robert Z
84.2k955123
edited Jul 17 at 13:45
Robert Z
84.2k955123
edited Jul 17 at 13:45
Robert Z
84.2k955123
84.2k955123
asked Jul 17 at 13:12
user515933
846
asked Jul 17 at 13:12
user515933
846
asked Jul 17 at 13:12
user515933
846
846
The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04
add a comment |Â
The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04
The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
HINT
Assuming that $f(x)$ is not constant (trivial case), by IVT and EVT we can show that $f(x)$ has maximum or a minimum for some $x_0in (a,infty)$ and then $f'(x_0)=0$.
answered Jul 17 at 13:19
gimusi
65.4k73584
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Assume $f'(x) not =0$ for $x>a$.
Then $f' >0$, or $f' <0$ for $x >a$.
1) Consider $f' >0$:
Then $f$ is strictly monotonically increasing.
For $x_1 >a$ , we have $f(x_1) >f(a)$.
For $x >x_1$ : $f(x) >f(x_1)>f(a).$
We are given:
$lim_x rightarrow infty f(x)=f(a)$, i.e.
for $epsilon >0$:
There is $K$, real, positive, such that for $x > K$
$|f(x)-f(a)| lt epsilon$, or
$ -epsilon + f(a) < f(x) < f(a) +epsilon.$
Choose $epsilon =f(x_1) - f(a)$ $(>0)$.
Let $K':= max (K,x_1)$,
for $x gt K' :$
$f(x)<f(a)+epsilon = f(x_1)$ ,
a contradiction.
Hence $f'(x) le 0$ for a $x >a$.
1) If $f'(x) = 0$ for a $x >a$ , we are finished.
2) If $f'(x) <0$ for a $x >a$, need continuity(??) of the derivative to infer $f'(x) = 0$ for a $x >a.$
answered Jul 17 at 15:51
Peter Szilas
7,9452617
add a comment |Â
up vote
0
down vote
For non-constant $f$.
(1). Suppose there exists $b>a$ such that $f(b)>f(a).$ Since f is continuous on $[a,b]$ there exists $cin (a,b)$ with $f(c)=(f(a)+f(b))/2.$
Since $lim_xto inftyf(x)=f(a)$ there exists $d>b$ with $f(d)<f(a)+delta ,$ where $delta =(f(b)-f(a))/2.$ That is, $f(d)<(f(a)+f(b))/2.$
Since $f$ is continuous on $[b,d]$ and $f(b)>(f(a)+f(b))/2>f(d),$ there exists $ein (b,d)$ with $f(e)=(f(a)+f(b))/2.$
We now have $a<c<e$ and $f(c)=f(e),$ and since f is differentiable on $(a,infty),$ there exists (by Rolle's Theorem) $x_0in (c,e)$ with $f'(x_0)=0.$
(2). If $f(b)< f(a)$ for some $b>a$ then this is handled similarly to (1) and I will leave the details to you.
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
HINT
Assuming that $f(x)$ is not constant (trivial case), by IVT and EVT we can show that $f(x)$ has maximum or a minimum for some $x_0in (a,infty)$ and then $f'(x_0)=0$.
answered Jul 17 at 13:19
gimusi
65.4k73584
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
add a comment |Â
up vote
2
down vote
HINT
Assuming that $f(x)$ is not constant (trivial case), by IVT and EVT we can show that $f(x)$ has maximum or a minimum for some $x_0in (a,infty)$ and then $f'(x_0)=0$.
answered Jul 17 at 13:19
gimusi
65.4k73584
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT
Assuming that $f(x)$ is not constant (trivial case), by IVT and EVT we can show that $f(x)$ has maximum or a minimum for some $x_0in (a,infty)$ and then $f'(x_0)=0$.
answered Jul 17 at 13:19
gimusi
65.4k73584
HINT
Assuming that $f(x)$ is not constant (trivial case), by IVT and EVT we can show that $f(x)$ has maximum or a minimum for some $x_0in (a,infty)$ and then $f'(x_0)=0$.
answered Jul 17 at 13:19
gimusi
65.4k73584
answered Jul 17 at 13:19
gimusi
65.4k73584
answered Jul 17 at 13:19
gimusi
65.4k73584
answered Jul 17 at 13:19
gimusi
65.4k73584
65.4k73584
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
add a comment |Â
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
But how do you use the hypothesis of the function limit?
– user515933
Jul 17 at 15:19
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
@user515933 If the interval of definition was a compact we could directly conclude by EVT but it is not the case. Then, assuming $f(x)$ is not constant, exists some $bin (a,infty)$ such that $f(b)neq f(a)$. Assume wlog $f(b)>f(a)$ and take $f(a)<f(c)<f(b)$ then by IVT exist $x_1in (a,b)$ and $x_2in (b,infty)$ such that $f(x_1)=f(x_2)=f(c)$ and by EVT for the restriction of the function onto $[x_1, x_2]$ exists a maximum at a point $x_0in (x_1, x_2)$.
– gimusi
Jul 17 at 15:35
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Assume $f'(x) not =0$ for $x>a$.
Then $f' >0$, or $f' <0$ for $x >a$.
1) Consider $f' >0$:
Then $f$ is strictly monotonically increasing.
For $x_1 >a$ , we have $f(x_1) >f(a)$.
For $x >x_1$ : $f(x) >f(x_1)>f(a).$
We are given:
$lim_x rightarrow infty f(x)=f(a)$, i.e.
for $epsilon >0$:
There is $K$, real, positive, such that for $x > K$
$|f(x)-f(a)| lt epsilon$, or
$ -epsilon + f(a) < f(x) < f(a) +epsilon.$
Choose $epsilon =f(x_1) - f(a)$ $(>0)$.
Let $K':= max (K,x_1)$,
for $x gt K' :$
$f(x)<f(a)+epsilon = f(x_1)$ ,
a contradiction.
Hence $f'(x) le 0$ for a $x >a$.
1) If $f'(x) = 0$ for a $x >a$ , we are finished.
2) If $f'(x) <0$ for a $x >a$, need continuity(??) of the derivative to infer $f'(x) = 0$ for a $x >a.$
answered Jul 17 at 15:51
Peter Szilas
7,9452617
add a comment |Â
up vote
0
down vote
Correct me if wrong.
Assume $f'(x) not =0$ for $x>a$.
Then $f' >0$, or $f' <0$ for $x >a$.
1) Consider $f' >0$:
Then $f$ is strictly monotonically increasing.
For $x_1 >a$ , we have $f(x_1) >f(a)$.
For $x >x_1$ : $f(x) >f(x_1)>f(a).$
We are given:
$lim_x rightarrow infty f(x)=f(a)$, i.e.
for $epsilon >0$:
There is $K$, real, positive, such that for $x > K$
$|f(x)-f(a)| lt epsilon$, or
$ -epsilon + f(a) < f(x) < f(a) +epsilon.$
Choose $epsilon =f(x_1) - f(a)$ $(>0)$.
Let $K':= max (K,x_1)$,
for $x gt K' :$
$f(x)<f(a)+epsilon = f(x_1)$ ,
a contradiction.
Hence $f'(x) le 0$ for a $x >a$.
1) If $f'(x) = 0$ for a $x >a$ , we are finished.
2) If $f'(x) <0$ for a $x >a$, need continuity(??) of the derivative to infer $f'(x) = 0$ for a $x >a.$
answered Jul 17 at 15:51
Peter Szilas
7,9452617
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Correct me if wrong.
Assume $f'(x) not =0$ for $x>a$.
Then $f' >0$, or $f' <0$ for $x >a$.
1) Consider $f' >0$:
Then $f$ is strictly monotonically increasing.
For $x_1 >a$ , we have $f(x_1) >f(a)$.
For $x >x_1$ : $f(x) >f(x_1)>f(a).$
We are given:
$lim_x rightarrow infty f(x)=f(a)$, i.e.
for $epsilon >0$:
There is $K$, real, positive, such that for $x > K$
$|f(x)-f(a)| lt epsilon$, or
$ -epsilon + f(a) < f(x) < f(a) +epsilon.$
Choose $epsilon =f(x_1) - f(a)$ $(>0)$.
Let $K':= max (K,x_1)$,
for $x gt K' :$
$f(x)<f(a)+epsilon = f(x_1)$ ,
a contradiction.
Hence $f'(x) le 0$ for a $x >a$.
1) If $f'(x) = 0$ for a $x >a$ , we are finished.
2) If $f'(x) <0$ for a $x >a$, need continuity(??) of the derivative to infer $f'(x) = 0$ for a $x >a.$
answered Jul 17 at 15:51
Peter Szilas
7,9452617
Correct me if wrong.
Assume $f'(x) not =0$ for $x>a$.
Then $f' >0$, or $f' <0$ for $x >a$.
1) Consider $f' >0$:
Then $f$ is strictly monotonically increasing.
For $x_1 >a$ , we have $f(x_1) >f(a)$.
For $x >x_1$ : $f(x) >f(x_1)>f(a).$
We are given:
$lim_x rightarrow infty f(x)=f(a)$, i.e.
for $epsilon >0$:
There is $K$, real, positive, such that for $x > K$
$|f(x)-f(a)| lt epsilon$, or
$ -epsilon + f(a) < f(x) < f(a) +epsilon.$
Choose $epsilon =f(x_1) - f(a)$ $(>0)$.
Let $K':= max (K,x_1)$,
for $x gt K' :$
$f(x)<f(a)+epsilon = f(x_1)$ ,
a contradiction.
Hence $f'(x) le 0$ for a $x >a$.
1) If $f'(x) = 0$ for a $x >a$ , we are finished.
2) If $f'(x) <0$ for a $x >a$, need continuity(??) of the derivative to infer $f'(x) = 0$ for a $x >a.$
answered Jul 17 at 15:51
Peter Szilas
7,9452617
answered Jul 17 at 15:51
Peter Szilas
7,9452617
answered Jul 17 at 15:51
Peter Szilas
7,9452617
answered Jul 17 at 15:51
Peter Szilas
7,9452617
7,9452617
add a comment |Â
add a comment |Â
up vote
0
down vote
For non-constant $f$.
(1). Suppose there exists $b>a$ such that $f(b)>f(a).$ Since f is continuous on $[a,b]$ there exists $cin (a,b)$ with $f(c)=(f(a)+f(b))/2.$
Since $lim_xto inftyf(x)=f(a)$ there exists $d>b$ with $f(d)<f(a)+delta ,$ where $delta =(f(b)-f(a))/2.$ That is, $f(d)<(f(a)+f(b))/2.$
Since $f$ is continuous on $[b,d]$ and $f(b)>(f(a)+f(b))/2>f(d),$ there exists $ein (b,d)$ with $f(e)=(f(a)+f(b))/2.$
We now have $a<c<e$ and $f(c)=f(e),$ and since f is differentiable on $(a,infty),$ there exists (by Rolle's Theorem) $x_0in (c,e)$ with $f'(x_0)=0.$
(2). If $f(b)< f(a)$ for some $b>a$ then this is handled similarly to (1) and I will leave the details to you.
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
add a comment |Â
up vote
0
down vote
For non-constant $f$.
(1). Suppose there exists $b>a$ such that $f(b)>f(a).$ Since f is continuous on $[a,b]$ there exists $cin (a,b)$ with $f(c)=(f(a)+f(b))/2.$
Since $lim_xto inftyf(x)=f(a)$ there exists $d>b$ with $f(d)<f(a)+delta ,$ where $delta =(f(b)-f(a))/2.$ That is, $f(d)<(f(a)+f(b))/2.$
Since $f$ is continuous on $[b,d]$ and $f(b)>(f(a)+f(b))/2>f(d),$ there exists $ein (b,d)$ with $f(e)=(f(a)+f(b))/2.$
We now have $a<c<e$ and $f(c)=f(e),$ and since f is differentiable on $(a,infty),$ there exists (by Rolle's Theorem) $x_0in (c,e)$ with $f'(x_0)=0.$
(2). If $f(b)< f(a)$ for some $b>a$ then this is handled similarly to (1) and I will leave the details to you.
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
add a comment |Â
up vote
0
down vote
up vote
0
down vote
For non-constant $f$.
(1). Suppose there exists $b>a$ such that $f(b)>f(a).$ Since f is continuous on $[a,b]$ there exists $cin (a,b)$ with $f(c)=(f(a)+f(b))/2.$
Since $lim_xto inftyf(x)=f(a)$ there exists $d>b$ with $f(d)<f(a)+delta ,$ where $delta =(f(b)-f(a))/2.$ That is, $f(d)<(f(a)+f(b))/2.$
Since $f$ is continuous on $[b,d]$ and $f(b)>(f(a)+f(b))/2>f(d),$ there exists $ein (b,d)$ with $f(e)=(f(a)+f(b))/2.$
We now have $a<c<e$ and $f(c)=f(e),$ and since f is differentiable on $(a,infty),$ there exists (by Rolle's Theorem) $x_0in (c,e)$ with $f'(x_0)=0.$
(2). If $f(b)< f(a)$ for some $b>a$ then this is handled similarly to (1) and I will leave the details to you.
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
For non-constant $f$.
(1). Suppose there exists $b>a$ such that $f(b)>f(a).$ Since f is continuous on $[a,b]$ there exists $cin (a,b)$ with $f(c)=(f(a)+f(b))/2.$
Since $lim_xto inftyf(x)=f(a)$ there exists $d>b$ with $f(d)<f(a)+delta ,$ where $delta =(f(b)-f(a))/2.$ That is, $f(d)<(f(a)+f(b))/2.$
Since $f$ is continuous on $[b,d]$ and $f(b)>(f(a)+f(b))/2>f(d),$ there exists $ein (b,d)$ with $f(e)=(f(a)+f(b))/2.$
We now have $a<c<e$ and $f(c)=f(e),$ and since f is differentiable on $(a,infty),$ there exists (by Rolle's Theorem) $x_0in (c,e)$ with $f'(x_0)=0.$
(2). If $f(b)< f(a)$ for some $b>a$ then this is handled similarly to (1) and I will leave the details to you.
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
answered Jul 17 at 18:30
DanielWainfleet
31.7k31644
31.7k31644
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
add a comment |Â
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
Intuitively, if $b>a$ and $f(b)ne f(a)$ then as $x$ moves from $b$ to $a,$ the value $f(x)$ moves continuously from $f(b)$ to $f(a),$ so $f$ has to take the value $(f(a)+f(b))/2$ at some point $cin (a,b).$.. And as $x$ moves from $b$ towards $infty,$ the value $f(x)$ moves continuously, starting at $f(b),$ and is eventually closer to $f(a)$ than to $(f(a)+f(b))/2 ,$ so $f$ has to take the value $(f(a)+f(b))/2$ at some at some point $ein (b,infty).$
– DanielWainfleet
Jul 17 at 18:45
add a comment |Â
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The grammatically correct expression in English is "$f$ is differentiable on $(a,infty)$".
– DanielWainfleet
Jul 17 at 18:50
@user515933 Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/…
– gimusi
Aug 8 at 23:04