Is this function continuous and differentiable?
$begingroup$
The function is $G(x,y)= 1$ if $y neq e^{x}$ and $0$ if $y= e^{x}$.
real-analysis calculus multivariable-calculus continuity differential
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
$endgroup$
add a comment |
$begingroup$
The function is $G(x,y)= 1$ if $y neq e^{x}$ and $0$ if $y= e^{x}$.
real-analysis calculus multivariable-calculus continuity differential
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
$endgroup$
add a comment |
$begingroup$
The function is $G(x,y)= 1$ if $y neq e^{x}$ and $0$ if $y= e^{x}$.
real-analysis calculus multivariable-calculus continuity differential
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
$endgroup$
The function is $G(x,y)= 1$ if $y neq e^{x}$ and $0$ if $y= e^{x}$.
real-analysis calculus multivariable-calculus continuity differential
real-analysis calculus multivariable-calculus continuity differential
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
edited Jan 22 at 20:43
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
asked Jan 22 at 19:20
M. NavarroM. Navarro
648
648
add a comment |
add a comment |
1 Answer
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$begingroup$
It is not continuous in the natural topology of $mathbb{R}^2$, because a continuous function cannot assume exactly two values.
Therefore it is not differentiable, either.
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
$endgroup$
add a comment |
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$begingroup$
It is not continuous in the natural topology of $mathbb{R}^2$, because a continuous function cannot assume exactly two values.
Therefore it is not differentiable, either.
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
$endgroup$
add a comment |
$begingroup$
It is not continuous in the natural topology of $mathbb{R}^2$, because a continuous function cannot assume exactly two values.
Therefore it is not differentiable, either.
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
$endgroup$
add a comment |
$begingroup$
It is not continuous in the natural topology of $mathbb{R}^2$, because a continuous function cannot assume exactly two values.
Therefore it is not differentiable, either.
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
$endgroup$
It is not continuous in the natural topology of $mathbb{R}^2$, because a continuous function cannot assume exactly two values.
Therefore it is not differentiable, either.
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
answered Jan 22 at 19:27
uniquesolutionuniquesolution
8,7711823
8,7711823
add a comment |
add a comment |
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