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Is it appropriate to consider a hole in the graph a zero?
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$begingroup$
Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?
functions terminology roots graphing-functions
asked yesterday
AnadactotheAnadactothe
13211
$endgroup$
add a comment |
$begingroup$
Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?
functions terminology roots graphing-functions
asked yesterday
AnadactotheAnadactothe
13211
$endgroup$
add a comment |
$begingroup$
Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?
functions terminology roots graphing-functions
asked yesterday
AnadactotheAnadactothe
13211
$endgroup$
Let there be a function $$f(x) = frac{x^2}{e^x-1}$$ which has a hole at $x=0$.
It also approaches $f(x) = 0$ at this point. Would it be appropriate to call this hole a "zero" or not?
functions terminology roots graphing-functions
functions terminology roots graphing-functions
asked yesterday
AnadactotheAnadactothe
13211
asked yesterday
AnadactotheAnadactothe
13211
asked yesterday
AnadactotheAnadactothe
13211
asked yesterday
AnadactotheAnadactothe
13211
asked yesterday
AnadactotheAnadactothe
13211
13211
add a comment |
add a comment |
1 Answer
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$begingroup$
A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.
On the other hand, since the discontinuity is removable, we can plug the hole by defining
$$g(x)=begin{cases}
f(x), &x neq 0\
0, &x=0
end{cases}$$
in which case $x=0$ is indeed a zero of $g(x)$.
answered yesterday
ThéophileThéophile
20.2k13047
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.
On the other hand, since the discontinuity is removable, we can plug the hole by defining
$$g(x)=begin{cases}
f(x), &x neq 0\
0, &x=0
end{cases}$$
in which case $x=0$ is indeed a zero of $g(x)$.
answered yesterday
ThéophileThéophile
20.2k13047
$endgroup$
add a comment |
$begingroup$
A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.
On the other hand, since the discontinuity is removable, we can plug the hole by defining
$$g(x)=begin{cases}
f(x), &x neq 0\
0, &x=0
end{cases}$$
in which case $x=0$ is indeed a zero of $g(x)$.
answered yesterday
ThéophileThéophile
20.2k13047
$endgroup$
add a comment |
$begingroup$
A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.
On the other hand, since the discontinuity is removable, we can plug the hole by defining
$$g(x)=begin{cases}
f(x), &x neq 0\
0, &x=0
end{cases}$$
in which case $x=0$ is indeed a zero of $g(x)$.
answered yesterday
ThéophileThéophile
20.2k13047
$endgroup$
A zero is a point $a$ where $f(a) = 0$. In your case, $f(0)$ is undefined, so no, $x=0$ is not a zero.
On the other hand, since the discontinuity is removable, we can plug the hole by defining
$$g(x)=begin{cases}
f(x), &x neq 0\
0, &x=0
end{cases}$$
in which case $x=0$ is indeed a zero of $g(x)$.
answered yesterday
ThéophileThéophile
20.2k13047
answered yesterday
ThéophileThéophile
20.2k13047
answered yesterday
ThéophileThéophile
20.2k13047
answered yesterday
ThéophileThéophile
20.2k13047
20.2k13047
add a comment |
add a comment |
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