Finding Constants for Continuity
Given the function $ f(x) = x^asin(frac{1}{x^b})$ if $x ne 0,$ $0$ when $x=0$
for which values of $a,b$ is $f'(x)$ continuous at $0$?
So the derivative is $f'(x)=ax^{a-1}sin(frac{1}{x^b})-bx^{a-b-1}cos(frac{1}{x^b})$.
Is this continuous for all $a,b in mathbb{R}$ at $x=0$? Since this is one of those oscillating functions, or because $f(x)$ is differentiable at $x=0$ only for $a > 2$, is that not true?
real-analysis
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
add a comment |
Given the function $ f(x) = x^asin(frac{1}{x^b})$ if $x ne 0,$ $0$ when $x=0$
for which values of $a,b$ is $f'(x)$ continuous at $0$?
So the derivative is $f'(x)=ax^{a-1}sin(frac{1}{x^b})-bx^{a-b-1}cos(frac{1}{x^b})$.
Is this continuous for all $a,b in mathbb{R}$ at $x=0$? Since this is one of those oscillating functions, or because $f(x)$ is differentiable at $x=0$ only for $a > 2$, is that not true?
real-analysis
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
1
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02
add a comment |
Given the function $ f(x) = x^asin(frac{1}{x^b})$ if $x ne 0,$ $0$ when $x=0$
for which values of $a,b$ is $f'(x)$ continuous at $0$?
So the derivative is $f'(x)=ax^{a-1}sin(frac{1}{x^b})-bx^{a-b-1}cos(frac{1}{x^b})$.
Is this continuous for all $a,b in mathbb{R}$ at $x=0$? Since this is one of those oscillating functions, or because $f(x)$ is differentiable at $x=0$ only for $a > 2$, is that not true?
real-analysis
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
Given the function $ f(x) = x^asin(frac{1}{x^b})$ if $x ne 0,$ $0$ when $x=0$
for which values of $a,b$ is $f'(x)$ continuous at $0$?
So the derivative is $f'(x)=ax^{a-1}sin(frac{1}{x^b})-bx^{a-b-1}cos(frac{1}{x^b})$.
Is this continuous for all $a,b in mathbb{R}$ at $x=0$? Since this is one of those oscillating functions, or because $f(x)$ is differentiable at $x=0$ only for $a > 2$, is that not true?
real-analysis
real-analysis
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
edited Nov 19 '18 at 2:59
Tianlalu
3,09621038
3,09621038
asked Nov 19 '18 at 2:39
clovis
459
asked Nov 19 '18 at 2:39
clovis
459
asked Nov 19 '18 at 2:39
clovis
459
459
Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
1
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02
add a comment |
Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
1
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02
Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
1
1
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02
add a comment |
1 Answer
1
active
oldest
votes
So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.
for $f'(0) = lim_{x->0} ax^{a-1}sin(frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.
and $f'(0) = lim_{x->0} bx^{a-b-1}cos(frac{1}{x^b})=0$
when $a-b-1 >0$ and doesn't exist otherwise.
answered Nov 19 '18 at 4:02
clovis
459
add a comment |
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1 Answer
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So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.
for $f'(0) = lim_{x->0} ax^{a-1}sin(frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.
and $f'(0) = lim_{x->0} bx^{a-b-1}cos(frac{1}{x^b})=0$
when $a-b-1 >0$ and doesn't exist otherwise.
answered Nov 19 '18 at 4:02
clovis
459
add a comment |
So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.
for $f'(0) = lim_{x->0} ax^{a-1}sin(frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.
and $f'(0) = lim_{x->0} bx^{a-b-1}cos(frac{1}{x^b})=0$
when $a-b-1 >0$ and doesn't exist otherwise.
answered Nov 19 '18 at 4:02
clovis
459
add a comment |
So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.
for $f'(0) = lim_{x->0} ax^{a-1}sin(frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.
and $f'(0) = lim_{x->0} bx^{a-b-1}cos(frac{1}{x^b})=0$
when $a-b-1 >0$ and doesn't exist otherwise.
answered Nov 19 '18 at 4:02
clovis
459
So, I used the definition of differentiability to see where the limit $f'(0)$ exists, and is therefore continuous.
for $f'(0) = lim_{x->0} ax^{a-1}sin(frac{1}{x^b})=0$ when $a-1 > 0, $ so $a>1$. b should be able to be any real number.
and $f'(0) = lim_{x->0} bx^{a-b-1}cos(frac{1}{x^b})=0$
when $a-b-1 >0$ and doesn't exist otherwise.
answered Nov 19 '18 at 4:02
clovis
459
answered Nov 19 '18 at 4:02
clovis
459
answered Nov 19 '18 at 4:02
clovis
459
answered Nov 19 '18 at 4:02
clovis
459
459
add a comment |
add a comment |
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Why won't you first evaluate $;f'(0);$ for whatever values of $;a,b;$ it exists?
– DonAntonio
Nov 19 '18 at 2:45
1
Check the link: math.stackexchange.com/questions/1175500/…
– Sujit Bhattacharyya
Nov 19 '18 at 3:02