Mixing
Continuity, differentiability of $f(x,y)=frac{x^{frac{3}{2}}y^b}{x^2+y^2}$ , $x^2+y^2ne 0 $
0
$begingroup$
$$f(x,y)=begin{cases}frac{x^{frac{3}{2}}y^b}{x^2+y^2}&,text{ when }x^2+y^2ne 0 \ 0&,text{ when }x^2+y^2=0end{cases}$$
Discuss $f$ ’s continuity, differentiability, partial derivative’s continuity.
My attempt:
Let $y=kx$ ,and $xrightarrow 0$
Then $f=(1+k^2)^{-1}x^{-frac{1}{2}+b}k^b$ then I guess I should be $>frac{1}{2}$ and $<$ and $=$
But I don’t know how to proof the continuity ,and differentiability ,and partial derivative’s continuity on the three situation
real-analysis limits multivariable-calculus derivatives continuity
edited Jan 8 at 11:18
StubbornAtom
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
$endgroup$
add a comment |
0
$begingroup$
$$f(x,y)=begin{cases}frac{x^{frac{3}{2}}y^b}{x^2+y^2}&,text{ when }x^2+y^2ne 0 \ 0&,text{ when }x^2+y^2=0end{cases}$$
Discuss $f$ ’s continuity, differentiability, partial derivative’s continuity.
My attempt:
Let $y=kx$ ,and $xrightarrow 0$
Then $f=(1+k^2)^{-1}x^{-frac{1}{2}+b}k^b$ then I guess I should be $>frac{1}{2}$ and $<$ and $=$
But I don’t know how to proof the continuity ,and differentiability ,and partial derivative’s continuity on the three situation
real-analysis limits multivariable-calculus derivatives continuity
edited Jan 8 at 11:18
StubbornAtom
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
$endgroup$
$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14
add a comment |
0
0
0
$begingroup$
$$f(x,y)=begin{cases}frac{x^{frac{3}{2}}y^b}{x^2+y^2}&,text{ when }x^2+y^2ne 0 \ 0&,text{ when }x^2+y^2=0end{cases}$$
Discuss $f$ ’s continuity, differentiability, partial derivative’s continuity.
My attempt:
Let $y=kx$ ,and $xrightarrow 0$
Then $f=(1+k^2)^{-1}x^{-frac{1}{2}+b}k^b$ then I guess I should be $>frac{1}{2}$ and $<$ and $=$
But I don’t know how to proof the continuity ,and differentiability ,and partial derivative’s continuity on the three situation
real-analysis limits multivariable-calculus derivatives continuity
edited Jan 8 at 11:18
StubbornAtom
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
$endgroup$
$$f(x,y)=begin{cases}frac{x^{frac{3}{2}}y^b}{x^2+y^2}&,text{ when }x^2+y^2ne 0 \ 0&,text{ when }x^2+y^2=0end{cases}$$
Discuss $f$ ’s continuity, differentiability, partial derivative’s continuity.
My attempt:
Let $y=kx$ ,and $xrightarrow 0$
Then $f=(1+k^2)^{-1}x^{-frac{1}{2}+b}k^b$ then I guess I should be $>frac{1}{2}$ and $<$ and $=$
But I don’t know how to proof the continuity ,and differentiability ,and partial derivative’s continuity on the three situation
real-analysis limits multivariable-calculus derivatives continuity
real-analysis limits multivariable-calculus derivatives continuity
edited Jan 8 at 11:18
StubbornAtom
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
edited Jan 8 at 11:18
StubbornAtom
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
edited Jan 8 at 11:18
StubbornAtom
5,75611138
edited Jan 8 at 11:18
StubbornAtom
5,75611138
edited Jan 8 at 11:18
StubbornAtom
5,75611138
5,75611138
asked Jan 8 at 10:58
jacksonjackson
829
asked Jan 8 at 10:58
jacksonjackson
829
asked Jan 8 at 10:58
jacksonjackson
829
829
$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14
add a comment |
$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14
$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14
add a comment |
1 Answer
1
active
oldest
votes
0
$begingroup$
Hint: Use that $$frac{1}{x^2+y^2}le frac{1}{2|xy|}$$
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Hint: Use that $$frac{1}{x^2+y^2}le frac{1}{2|xy|}$$
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
add a comment |
0
$begingroup$
Hint: Use that $$frac{1}{x^2+y^2}le frac{1}{2|xy|}$$
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
add a comment |
0
0
0
$begingroup$
Hint: Use that $$frac{1}{x^2+y^2}le frac{1}{2|xy|}$$
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
$endgroup$
Hint: Use that $$frac{1}{x^2+y^2}le frac{1}{2|xy|}$$
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
answered Jan 8 at 11:24
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74.6k42865
74.6k42865
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
add a comment |
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
So when b$ge 1 is continuity ,but when b<1 what should I do
$endgroup$
– jackson
Jan 8 at 11:32
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
In this case we have no continuity
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:36
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
how I proof it I
$endgroup$
– jackson
Jan 8 at 11:42
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
In this case $$|f(x,y)|$$ tends not to zero
$endgroup$
– Dr. Sonnhard Graubner
Jan 8 at 11:57
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
$begingroup$
but how I proof it I still don’t figure it out ,can you write explicit to me thanks very much
$endgroup$
– jackson
Jan 8 at 13:02
add a comment |
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$begingroup$
"Then I guess I should be > 1/2" is meaningless. Are we to presume f is a real value function?
$endgroup$
– William Elliot
Jan 8 at 11:25
$begingroup$
@WilliamElliot yes we assume f is real function
$endgroup$
– jackson
Jan 8 at 11:27
$begingroup$
Then you should restrict the domain of f to nonnegative numbers.
$endgroup$
– William Elliot
Jan 8 at 21:14