Reciprocal Rational Function
$begingroup$
Suppose we have a rational function:
$$f(x) = dfrac{a+x}{b + cx}$$
And our task is to draw the reciprocal function of $f(x)$, or in other words, $frac{1}{f(x)}$.
My teacher argues that because the point at $x = -2$ is a vertical asymptote, we can say that the y-coordinate at the point is $∞$, and thus that the that the reciprocal function would have a x-intercept at $x = -2$.
I say that because we say can say that the point at $x = -2$ is undefined, in the reciprocal function we would have the point $dfrac{1}{undefined}$, which is also undefined, and therefore, there would not be an x-intercept, but rather an open circle at $x = -2$.
We cannot say that the point (-2, $∞$) exists on $f(x)$, and this is another point to back my point.
Who is correct? I understand that it is the fact that undefined and infinity are used interchangeably to describe the y-coordinate at a vertical asymptote that is causing the confusion. Could anyone please clarify, as the two are used as if they are equal but aren't...
Any help will be greatly appreciated, thanks in advance.
rational-functions
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
$endgroup$
add a comment |
$begingroup$
Suppose we have a rational function:
$$f(x) = dfrac{a+x}{b + cx}$$
And our task is to draw the reciprocal function of $f(x)$, or in other words, $frac{1}{f(x)}$.
My teacher argues that because the point at $x = -2$ is a vertical asymptote, we can say that the y-coordinate at the point is $∞$, and thus that the that the reciprocal function would have a x-intercept at $x = -2$.
I say that because we say can say that the point at $x = -2$ is undefined, in the reciprocal function we would have the point $dfrac{1}{undefined}$, which is also undefined, and therefore, there would not be an x-intercept, but rather an open circle at $x = -2$.
We cannot say that the point (-2, $∞$) exists on $f(x)$, and this is another point to back my point.
Who is correct? I understand that it is the fact that undefined and infinity are used interchangeably to describe the y-coordinate at a vertical asymptote that is causing the confusion. Could anyone please clarify, as the two are used as if they are equal but aren't...
Any help will be greatly appreciated, thanks in advance.
rational-functions
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
$endgroup$
$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
add a comment |
$begingroup$
Suppose we have a rational function:
$$f(x) = dfrac{a+x}{b + cx}$$
And our task is to draw the reciprocal function of $f(x)$, or in other words, $frac{1}{f(x)}$.
My teacher argues that because the point at $x = -2$ is a vertical asymptote, we can say that the y-coordinate at the point is $∞$, and thus that the that the reciprocal function would have a x-intercept at $x = -2$.
I say that because we say can say that the point at $x = -2$ is undefined, in the reciprocal function we would have the point $dfrac{1}{undefined}$, which is also undefined, and therefore, there would not be an x-intercept, but rather an open circle at $x = -2$.
We cannot say that the point (-2, $∞$) exists on $f(x)$, and this is another point to back my point.
Who is correct? I understand that it is the fact that undefined and infinity are used interchangeably to describe the y-coordinate at a vertical asymptote that is causing the confusion. Could anyone please clarify, as the two are used as if they are equal but aren't...
Any help will be greatly appreciated, thanks in advance.
rational-functions
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
$endgroup$
Suppose we have a rational function:
$$f(x) = dfrac{a+x}{b + cx}$$
And our task is to draw the reciprocal function of $f(x)$, or in other words, $frac{1}{f(x)}$.
My teacher argues that because the point at $x = -2$ is a vertical asymptote, we can say that the y-coordinate at the point is $∞$, and thus that the that the reciprocal function would have a x-intercept at $x = -2$.
I say that because we say can say that the point at $x = -2$ is undefined, in the reciprocal function we would have the point $dfrac{1}{undefined}$, which is also undefined, and therefore, there would not be an x-intercept, but rather an open circle at $x = -2$.
We cannot say that the point (-2, $∞$) exists on $f(x)$, and this is another point to back my point.
Who is correct? I understand that it is the fact that undefined and infinity are used interchangeably to describe the y-coordinate at a vertical asymptote that is causing the confusion. Could anyone please clarify, as the two are used as if they are equal but aren't...
Any help will be greatly appreciated, thanks in advance.
rational-functions
rational-functions
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
edited Oct 18 '15 at 2:13
GoodChessPlayer
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
asked Oct 18 '15 at 1:55
GoodChessPlayerGoodChessPlayer
98110
98110
$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
add a comment |
$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It depends on whether you can simplify $frac{1}{1/0}$ as $0$ or not.
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
add a comment |
$begingroup$
The expression “rational function” may be somewhat ambiguous. Do we mean a function in the standard calculus sense, where it would be $f:(Bbb Rsetminus S)toBbb R$, where $S$ is a finite set, corresponding to the zeroes of the denominator in $f(x)=n(x)/d(x)$, where $n$ and $d$ are polynomials with real coefficients. In this case, $1/f$ certainly does fail to be defined at the zeros of $f$, the elements of $S$.
But in abstract algebra, a rational function $f(x)=n(x)/d(x)$ is slightly different: it’s just the formal expression that you see, a polynomial over another polynomial, and in this case the reciprocal is just $d/n$, another formal expression, which as a calculus-style function will fail to be defined at another finite set than $S$. You pays your money and takes your choice.
Others will have different ways to get around the difficulty between you and your teacher; I for one say both of you are right, according to your lights.
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It depends on whether you can simplify $frac{1}{1/0}$ as $0$ or not.
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
add a comment |
$begingroup$
It depends on whether you can simplify $frac{1}{1/0}$ as $0$ or not.
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
$endgroup$
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
add a comment |
$begingroup$
It depends on whether you can simplify $frac{1}{1/0}$ as $0$ or not.
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
$endgroup$
It depends on whether you can simplify $frac{1}{1/0}$ as $0$ or not.
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
answered Oct 18 '15 at 1:58
Henricus V.Henricus V.
15.1k22048
15.1k22048
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
add a comment |
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
$begingroup$
Can you please elaborate?
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:16
add a comment |
$begingroup$
The expression “rational function” may be somewhat ambiguous. Do we mean a function in the standard calculus sense, where it would be $f:(Bbb Rsetminus S)toBbb R$, where $S$ is a finite set, corresponding to the zeroes of the denominator in $f(x)=n(x)/d(x)$, where $n$ and $d$ are polynomials with real coefficients. In this case, $1/f$ certainly does fail to be defined at the zeros of $f$, the elements of $S$.
But in abstract algebra, a rational function $f(x)=n(x)/d(x)$ is slightly different: it’s just the formal expression that you see, a polynomial over another polynomial, and in this case the reciprocal is just $d/n$, another formal expression, which as a calculus-style function will fail to be defined at another finite set than $S$. You pays your money and takes your choice.
Others will have different ways to get around the difficulty between you and your teacher; I for one say both of you are right, according to your lights.
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
$endgroup$
add a comment |
$begingroup$
The expression “rational function” may be somewhat ambiguous. Do we mean a function in the standard calculus sense, where it would be $f:(Bbb Rsetminus S)toBbb R$, where $S$ is a finite set, corresponding to the zeroes of the denominator in $f(x)=n(x)/d(x)$, where $n$ and $d$ are polynomials with real coefficients. In this case, $1/f$ certainly does fail to be defined at the zeros of $f$, the elements of $S$.
But in abstract algebra, a rational function $f(x)=n(x)/d(x)$ is slightly different: it’s just the formal expression that you see, a polynomial over another polynomial, and in this case the reciprocal is just $d/n$, another formal expression, which as a calculus-style function will fail to be defined at another finite set than $S$. You pays your money and takes your choice.
Others will have different ways to get around the difficulty between you and your teacher; I for one say both of you are right, according to your lights.
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
$endgroup$
add a comment |
$begingroup$
The expression “rational function” may be somewhat ambiguous. Do we mean a function in the standard calculus sense, where it would be $f:(Bbb Rsetminus S)toBbb R$, where $S$ is a finite set, corresponding to the zeroes of the denominator in $f(x)=n(x)/d(x)$, where $n$ and $d$ are polynomials with real coefficients. In this case, $1/f$ certainly does fail to be defined at the zeros of $f$, the elements of $S$.
But in abstract algebra, a rational function $f(x)=n(x)/d(x)$ is slightly different: it’s just the formal expression that you see, a polynomial over another polynomial, and in this case the reciprocal is just $d/n$, another formal expression, which as a calculus-style function will fail to be defined at another finite set than $S$. You pays your money and takes your choice.
Others will have different ways to get around the difficulty between you and your teacher; I for one say both of you are right, according to your lights.
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
$endgroup$
The expression “rational function” may be somewhat ambiguous. Do we mean a function in the standard calculus sense, where it would be $f:(Bbb Rsetminus S)toBbb R$, where $S$ is a finite set, corresponding to the zeroes of the denominator in $f(x)=n(x)/d(x)$, where $n$ and $d$ are polynomials with real coefficients. In this case, $1/f$ certainly does fail to be defined at the zeros of $f$, the elements of $S$.
But in abstract algebra, a rational function $f(x)=n(x)/d(x)$ is slightly different: it’s just the formal expression that you see, a polynomial over another polynomial, and in this case the reciprocal is just $d/n$, another formal expression, which as a calculus-style function will fail to be defined at another finite set than $S$. You pays your money and takes your choice.
Others will have different ways to get around the difficulty between you and your teacher; I for one say both of you are right, according to your lights.
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
answered Oct 18 '15 at 2:30
LubinLubin
44.3k44585
44.3k44585
add a comment |
add a comment |
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$begingroup$
do you know what $a,b,c$ are?
$endgroup$
– Will Jagy
Oct 18 '15 at 2:06
$begingroup$
What exactly do you mean by reciprocating? Are we interchanging numerator and denominator or are we taking an inverse?
$endgroup$
– imranfat
Oct 18 '15 at 2:08
$begingroup$
Yes, it is something you can deduct from the given graph. $a = -2$, $b = 1$, $c = frac{1}{2}$.
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10
$begingroup$
No, not an inverse. We are trying to find the graph for $frac{1}{f(x)}$
$endgroup$
– GoodChessPlayer
Oct 18 '15 at 2:10