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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
edited Jul 20 at 17:17
Ethan Bolker
35.7k54199
asked Jul 20 at 16:54
LOIS
947
add a comment |Â
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
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 |Â
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
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
edited Jul 20 at 17:17
Ethan Bolker
35.7k54199
asked Jul 20 at 16:54
LOIS
947
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 |Â
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 |Â
1 Answer
1
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4
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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.
answered Jul 20 at 17:24
Raymond Chu
1,03719
add a comment |Â
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.
answered Jul 20 at 17:24
Raymond Chu
1,03719
add a comment |Â
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.
answered Jul 20 at 17:24
Raymond Chu
1,03719
add a comment |Â
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.
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.
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
answered Jul 20 at 17:24
Raymond Chu
1,03719
1,03719
add a comment |Â
add a comment |Â
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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