Assume $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$.

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Assume $f$ is continuous on $[a,b]$ and differientiable on $(a,b)$. Also $lim_xto a^+f'(x) $ exists. Then the question is whether $f$ has a right derivative at point $x=a$.

Here is my tries.

$$lim_xto a^+fracf'(x)-f'(a)x-a=lim_tto 0^+fracf'(a+t)-f'(a)t=k $$ for $lim_xto a^+f'(x) $ exists.

Then consider
$$lim_tto 0^+ fracf(a+t)-f(a)t=lim_tto 0^+ f'(xi)=k$$
which $xi in(a,a+t)$.

Now we can conclude $f$ has a right derivative at point $x=a$.

Am I right?

calculus

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edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

asked Jul 20 at 16:54

LOIS

947

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As with just about any basic question about derivatives: Mean Value Theorem.
  – David C. Ullrich
  Jul 20 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur Sorry,I have edited it. Please have a look.
  – LOIS
  Jul 20 at 17:17
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

Assume $f$ is continuous on $[a,b]$ and differientiable on $(a,b)$. Also $lim_xto a^+f'(x) $ exists. Then the question is whether $f$ has a right derivative at point $x=a$.

Here is my tries.

$$lim_xto a^+fracf'(x)-f'(a)x-a=lim_tto 0^+fracf'(a+t)-f'(a)t=k $$ for $lim_xto a^+f'(x) $ exists.

Then consider
$$lim_tto 0^+ fracf(a+t)-f(a)t=lim_tto 0^+ f'(xi)=k$$
which $xi in(a,a+t)$.

Now we can conclude $f$ has a right derivative at point $x=a$.

Am I right?

calculus

share|cite|improve this question

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

asked Jul 20 at 16:54

LOIS

947

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As with just about any basic question about derivatives: Mean Value Theorem.
  – David C. Ullrich
  Jul 20 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur Sorry,I have edited it. Please have a look.
  – LOIS
  Jul 20 at 17:17
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Assume $f$ is continuous on $[a,b]$ and differientiable on $(a,b)$. Also $lim_xto a^+f'(x) $ exists. Then the question is whether $f$ has a right derivative at point $x=a$.

Here is my tries.

$$lim_xto a^+fracf'(x)-f'(a)x-a=lim_tto 0^+fracf'(a+t)-f'(a)t=k $$ for $lim_xto a^+f'(x) $ exists.

Then consider
$$lim_tto 0^+ fracf(a+t)-f(a)t=lim_tto 0^+ f'(xi)=k$$
which $xi in(a,a+t)$.

Now we can conclude $f$ has a right derivative at point $x=a$.

Am I right?

calculus

share|cite|improve this question

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

asked Jul 20 at 16:54

LOIS

947

Assume $f$ is continuous on $[a,b]$ and differientiable on $(a,b)$. Also $lim_xto a^+f'(x) $ exists. Then the question is whether $f$ has a right derivative at point $x=a$.

Here is my tries.

$$lim_xto a^+fracf'(x)-f'(a)x-a=lim_tto 0^+fracf'(a+t)-f'(a)t=k $$ for $lim_xto a^+f'(x) $ exists.

Then consider
$$lim_tto 0^+ fracf(a+t)-f(a)t=lim_tto 0^+ f'(xi)=k$$
which $xi in(a,a+t)$.

Now we can conclude $f$ has a right derivative at point $x=a$.

Am I right?

calculus

share|cite|improve this question

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

asked Jul 20 at 16:54

LOIS

947

share|cite|improve this question

share|cite|improve this question

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

edited Jul 20 at 17:17

Ethan Bolker

35.7k54199

35.7k54199

asked Jul 20 at 16:54

LOIS

947

asked Jul 20 at 16:54

LOIS

947

asked Jul 20 at 16:54

LOIS

947

947

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As with just about any basic question about derivatives: Mean Value Theorem.
  – David C. Ullrich
  Jul 20 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur Sorry,I have edited it. Please have a look.
  – LOIS
  Jul 20 at 17:17
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  As with just about any basic question about derivatives: Mean Value Theorem.
  – David C. Ullrich
  Jul 20 at 17:04
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur Sorry,I have edited it. Please have a look.
  – LOIS
  Jul 20 at 17:17
  

  
  
  
  

As with just about any basic question about derivatives: Mean Value Theorem.
– David C. Ullrich
Jul 20 at 17:04

As with just about any basic question about derivatives: Mean Value Theorem.
– David C. Ullrich
Jul 20 at 17:04

@Arthur Sorry,I have edited it. Please have a look.
– LOIS
Jul 20 at 17:17

@Arthur Sorry,I have edited it. Please have a look.
– LOIS
Jul 20 at 17:17

add a comment | 

1 Answer
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This is a basic question on MVT.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that $
lim_x rightarrow a^+f'(x)$ exists. As the limit exists, for every sequence that goes downward to $a$, $f'$ converges to that value.

Then we see on the interval $(a,b)$ for any sequence $h_n$ such that $h_n rightarrow 0$ after some $N$, we have that $(a,a+h_n) subset (a,b)$

Consider $a + h_k_k=n^infty$. Then by MVT there is some $zeta in (a,a+h_k)$ such that beginalign fracf(a+h_k)-f(a)h_k = f'(zeta) endalign Then we see as $k rightarrow infty$ beginalign lim_k rightarrow infty fracf(a+h_k)-f(a)h_k rightarrow lim_x rightarrow a^+ f'(x) endalign
where this is justified by the limit existing.And as this was an arbitrary subsequence that converges to $a$, we obtain the claim.

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

This is a basic question on MVT.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that $
lim_x rightarrow a^+f'(x)$ exists. As the limit exists, for every sequence that goes downward to $a$, $f'$ converges to that value.

Then we see on the interval $(a,b)$ for any sequence $h_n$ such that $h_n rightarrow 0$ after some $N$, we have that $(a,a+h_n) subset (a,b)$

Consider $a + h_k_k=n^infty$. Then by MVT there is some $zeta in (a,a+h_k)$ such that beginalign fracf(a+h_k)-f(a)h_k = f'(zeta) endalign Then we see as $k rightarrow infty$ beginalign lim_k rightarrow infty fracf(a+h_k)-f(a)h_k rightarrow lim_x rightarrow a^+ f'(x) endalign
where this is justified by the limit existing.And as this was an arbitrary subsequence that converges to $a$, we obtain the claim.

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

add a comment | 

up vote
4
down vote

This is a basic question on MVT.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that $
lim_x rightarrow a^+f'(x)$ exists. As the limit exists, for every sequence that goes downward to $a$, $f'$ converges to that value.

Then we see on the interval $(a,b)$ for any sequence $h_n$ such that $h_n rightarrow 0$ after some $N$, we have that $(a,a+h_n) subset (a,b)$

Consider $a + h_k_k=n^infty$. Then by MVT there is some $zeta in (a,a+h_k)$ such that beginalign fracf(a+h_k)-f(a)h_k = f'(zeta) endalign Then we see as $k rightarrow infty$ beginalign lim_k rightarrow infty fracf(a+h_k)-f(a)h_k rightarrow lim_x rightarrow a^+ f'(x) endalign
where this is justified by the limit existing.And as this was an arbitrary subsequence that converges to $a$, we obtain the claim.

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

add a comment | 

up vote
4
down vote

up vote
4
down vote

This is a basic question on MVT.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that $
lim_x rightarrow a^+f'(x)$ exists. As the limit exists, for every sequence that goes downward to $a$, $f'$ converges to that value.

Then we see on the interval $(a,b)$ for any sequence $h_n$ such that $h_n rightarrow 0$ after some $N$, we have that $(a,a+h_n) subset (a,b)$

Consider $a + h_k_k=n^infty$. Then by MVT there is some $zeta in (a,a+h_k)$ such that beginalign fracf(a+h_k)-f(a)h_k = f'(zeta) endalign Then we see as $k rightarrow infty$ beginalign lim_k rightarrow infty fracf(a+h_k)-f(a)h_k rightarrow lim_x rightarrow a^+ f'(x) endalign
where this is justified by the limit existing.And as this was an arbitrary subsequence that converges to $a$, we obtain the claim.

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

This is a basic question on MVT.

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ such that $
lim_x rightarrow a^+f'(x)$ exists. As the limit exists, for every sequence that goes downward to $a$, $f'$ converges to that value.

Then we see on the interval $(a,b)$ for any sequence $h_n$ such that $h_n rightarrow 0$ after some $N$, we have that $(a,a+h_n) subset (a,b)$

Consider $a + h_k_k=n^infty$. Then by MVT there is some $zeta in (a,a+h_k)$ such that beginalign fracf(a+h_k)-f(a)h_k = f'(zeta) endalign Then we see as $k rightarrow infty$ beginalign lim_k rightarrow infty fracf(a+h_k)-f(a)h_k rightarrow lim_x rightarrow a^+ f'(x) endalign
where this is justified by the limit existing.And as this was an arbitrary subsequence that converges to $a$, we obtain the claim.

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 17:24

Raymond Chu

1,03719

answered Jul 20 at 17:24

Raymond Chu

1,03719

answered Jul 20 at 17:24

Raymond Chu

1,03719

1,03719

add a comment | 

add a comment | 

 

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