Mixing
Why are these functions not equal?
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Fairly straightforward question: $$gleft(xright)=fracxx+1$$ $$hleft(xright)=fracx^2x^2+x$$
Both $g$ and $h$ have the same graph, the same domain and range. So why did I read that these are not equal?
functions
asked 22 mins ago
Daniel Oscar
825
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up vote
4
down vote
favorite
Fairly straightforward question: $$gleft(xright)=fracxx+1$$ $$hleft(xright)=fracx^2x^2+x$$
Both $g$ and $h$ have the same graph, the same domain and range. So why did I read that these are not equal?
functions
asked 22 mins ago
Daniel Oscar
825
1
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Fairly straightforward question: $$gleft(xright)=fracxx+1$$ $$hleft(xright)=fracx^2x^2+x$$
Both $g$ and $h$ have the same graph, the same domain and range. So why did I read that these are not equal?
functions
asked 22 mins ago
Daniel Oscar
825
Fairly straightforward question: $$gleft(xright)=fracxx+1$$ $$hleft(xright)=fracx^2x^2+x$$
Both $g$ and $h$ have the same graph, the same domain and range. So why did I read that these are not equal?
functions
functions
asked 22 mins ago
Daniel Oscar
825
asked 22 mins ago
Daniel Oscar
825
asked 22 mins ago
Daniel Oscar
825
asked 22 mins ago
Daniel Oscar
825
asked 22 mins ago
Daniel Oscar
825
825
1
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago
add a comment |Â
1
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago
1
1
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
$g(0)$ exists while $h(0)$ is undefined (because in the latter function, you would be dividing by zero).
answered 21 mins ago
InterstellarProbe
3,156724
add a comment |Â
up vote
1
down vote
Note that
$g(x)$ is not defined at $x=-1$
$f(x)$ is not defined at $x=-1$ and $x=0$
therefore the two functions are equal for any $xneq 0$ and they become equal if we define $f(0)=0$ for $x=0$ which is apoint of removable discontinuity for $f(x)$.
answered 21 mins ago
gimusi
74.8k73889
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
add a comment |Â
up vote
1
down vote
Both $g$ and $h$ have the same graph, the same domain and range.
They don't have the same domain: $g$ is not defined at $x=-1$, while $h$ is not defined at $x=0,-1$. You can't ignore that point unless you're considering another (smaller) domain not containing $x=0$. Then $g=h$.
Indeed,
$$fracx^2x^2+x=fracx1+x iff xneq0$$
answered 11 mins ago
edmz
431311
@amWhy: Surely, thanks.
– edmz
7 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$g(0)$ exists while $h(0)$ is undefined (because in the latter function, you would be dividing by zero).
answered 21 mins ago
InterstellarProbe
3,156724
add a comment |Â
up vote
3
down vote
$g(0)$ exists while $h(0)$ is undefined (because in the latter function, you would be dividing by zero).
answered 21 mins ago
InterstellarProbe
3,156724
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$g(0)$ exists while $h(0)$ is undefined (because in the latter function, you would be dividing by zero).
answered 21 mins ago
InterstellarProbe
3,156724
$g(0)$ exists while $h(0)$ is undefined (because in the latter function, you would be dividing by zero).
answered 21 mins ago
InterstellarProbe
3,156724
answered 21 mins ago
InterstellarProbe
3,156724
answered 21 mins ago
InterstellarProbe
3,156724
answered 21 mins ago
InterstellarProbe
3,156724
3,156724
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that
$g(x)$ is not defined at $x=-1$
$f(x)$ is not defined at $x=-1$ and $x=0$
therefore the two functions are equal for any $xneq 0$ and they become equal if we define $f(0)=0$ for $x=0$ which is apoint of removable discontinuity for $f(x)$.
answered 21 mins ago
gimusi
74.8k73889
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
add a comment |Â
up vote
1
down vote
Note that
$g(x)$ is not defined at $x=-1$
$f(x)$ is not defined at $x=-1$ and $x=0$
therefore the two functions are equal for any $xneq 0$ and they become equal if we define $f(0)=0$ for $x=0$ which is apoint of removable discontinuity for $f(x)$.
answered 21 mins ago
gimusi
74.8k73889
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that
$g(x)$ is not defined at $x=-1$
$f(x)$ is not defined at $x=-1$ and $x=0$
therefore the two functions are equal for any $xneq 0$ and they become equal if we define $f(0)=0$ for $x=0$ which is apoint of removable discontinuity for $f(x)$.
answered 21 mins ago
gimusi
74.8k73889
Note that
$g(x)$ is not defined at $x=-1$
$f(x)$ is not defined at $x=-1$ and $x=0$
therefore the two functions are equal for any $xneq 0$ and they become equal if we define $f(0)=0$ for $x=0$ which is apoint of removable discontinuity for $f(x)$.
answered 21 mins ago
gimusi
74.8k73889
edited 14 mins ago
answered 21 mins ago
gimusi
74.8k73889
answered 21 mins ago
gimusi
74.8k73889
answered 21 mins ago
gimusi
74.8k73889
74.8k73889
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
add a comment |Â
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
Right! I guess I could've figured that out, my bad! It seems I was induced in error when I plotted both functions in geogebra. But then why do the graphs seem equal? I'm guessing that if I could zoom to infinity I would see a discontinuity at the points of non-definition?
– Daniel Oscar
12 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
@DanielOscar Matbe not by the graph but by the expression we can easily check for that!
– gimusi
10 mins ago
add a comment |Â
up vote
1
down vote
Both $g$ and $h$ have the same graph, the same domain and range.
They don't have the same domain: $g$ is not defined at $x=-1$, while $h$ is not defined at $x=0,-1$. You can't ignore that point unless you're considering another (smaller) domain not containing $x=0$. Then $g=h$.
Indeed,
$$fracx^2x^2+x=fracx1+x iff xneq0$$
answered 11 mins ago
edmz
431311
@amWhy: Surely, thanks.
– edmz
7 mins ago
add a comment |Â
up vote
1
down vote
Both $g$ and $h$ have the same graph, the same domain and range.
They don't have the same domain: $g$ is not defined at $x=-1$, while $h$ is not defined at $x=0,-1$. You can't ignore that point unless you're considering another (smaller) domain not containing $x=0$. Then $g=h$.
Indeed,
$$fracx^2x^2+x=fracx1+x iff xneq0$$
answered 11 mins ago
edmz
431311
@amWhy: Surely, thanks.
– edmz
7 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Both $g$ and $h$ have the same graph, the same domain and range.
They don't have the same domain: $g$ is not defined at $x=-1$, while $h$ is not defined at $x=0,-1$. You can't ignore that point unless you're considering another (smaller) domain not containing $x=0$. Then $g=h$.
Indeed,
$$fracx^2x^2+x=fracx1+x iff xneq0$$
answered 11 mins ago
edmz
431311
Both $g$ and $h$ have the same graph, the same domain and range.
They don't have the same domain: $g$ is not defined at $x=-1$, while $h$ is not defined at $x=0,-1$. You can't ignore that point unless you're considering another (smaller) domain not containing $x=0$. Then $g=h$.
Indeed,
$$fracx^2x^2+x=fracx1+x iff xneq0$$
answered 11 mins ago
edmz
431311
edited 7 mins ago
answered 11 mins ago
edmz
431311
answered 11 mins ago
edmz
431311
answered 11 mins ago
edmz
431311
431311
@amWhy: Surely, thanks.
– edmz
7 mins ago
add a comment |Â
@amWhy: Surely, thanks.
– edmz
7 mins ago
@amWhy: Surely, thanks.
– edmz
7 mins ago
@amWhy: Surely, thanks.
– edmz
7 mins ago
add a comment |Â
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1
The graphs are not the same. The graph of $f(x)$ has a removable discontinuity at $x=0$
– Zubin Mukerjee
19 mins ago