Why the Graph of the function $f(x)=frac{x^2-3x+2}{x^2+2x-3}$ has not excluded $frac{-1}{4}$ from the Range
$begingroup$
I was trying to find range of the function :
$$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$
we have the Domain as:
$$D_f=mathbb{R}-left{-3,1right}$$
We have $$y=f(x)=frac{(x-1)(x-2)}{(x+3)(x-1)}=frac{x-2}{x+3}tag{1}$$
So we get:
$$x=frac{3y+2}{1-y}$$
so $y ne 1$ Also when $x=1$ in Equation $(1)$ we get $y=frac{-1}{4}$
Since $x=1$ is not in domain we exclude $y=frac{-1}{4}$ in range.
hence range is $$D_r=mathbb{R}-left{1,frac{-1}{4}right}$$
But when I draw the graph of $$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$ in Desmos, why the value $frac{-1}{4}$ is not excluded in range?
algebra-precalculus polynomials graphing-functions
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
$endgroup$
add a comment |
$begingroup$
I was trying to find range of the function :
$$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$
we have the Domain as:
$$D_f=mathbb{R}-left{-3,1right}$$
We have $$y=f(x)=frac{(x-1)(x-2)}{(x+3)(x-1)}=frac{x-2}{x+3}tag{1}$$
So we get:
$$x=frac{3y+2}{1-y}$$
so $y ne 1$ Also when $x=1$ in Equation $(1)$ we get $y=frac{-1}{4}$
Since $x=1$ is not in domain we exclude $y=frac{-1}{4}$ in range.
hence range is $$D_r=mathbb{R}-left{1,frac{-1}{4}right}$$
But when I draw the graph of $$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$ in Desmos, why the value $frac{-1}{4}$ is not excluded in range?
algebra-precalculus polynomials graphing-functions
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
$endgroup$
3
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40
add a comment |
$begingroup$
I was trying to find range of the function :
$$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$
we have the Domain as:
$$D_f=mathbb{R}-left{-3,1right}$$
We have $$y=f(x)=frac{(x-1)(x-2)}{(x+3)(x-1)}=frac{x-2}{x+3}tag{1}$$
So we get:
$$x=frac{3y+2}{1-y}$$
so $y ne 1$ Also when $x=1$ in Equation $(1)$ we get $y=frac{-1}{4}$
Since $x=1$ is not in domain we exclude $y=frac{-1}{4}$ in range.
hence range is $$D_r=mathbb{R}-left{1,frac{-1}{4}right}$$
But when I draw the graph of $$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$ in Desmos, why the value $frac{-1}{4}$ is not excluded in range?
algebra-precalculus polynomials graphing-functions
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
$endgroup$
I was trying to find range of the function :
$$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$
we have the Domain as:
$$D_f=mathbb{R}-left{-3,1right}$$
We have $$y=f(x)=frac{(x-1)(x-2)}{(x+3)(x-1)}=frac{x-2}{x+3}tag{1}$$
So we get:
$$x=frac{3y+2}{1-y}$$
so $y ne 1$ Also when $x=1$ in Equation $(1)$ we get $y=frac{-1}{4}$
Since $x=1$ is not in domain we exclude $y=frac{-1}{4}$ in range.
hence range is $$D_r=mathbb{R}-left{1,frac{-1}{4}right}$$
But when I draw the graph of $$f(x)=frac{x^2-3x+2}{x^2+2x-3}$$ in Desmos, why the value $frac{-1}{4}$ is not excluded in range?
algebra-precalculus polynomials graphing-functions
algebra-precalculus polynomials graphing-functions
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
asked Dec 23 '18 at 5:16
Umesh shankarUmesh shankar
2,72231220
2,72231220
3
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40
add a comment |
3
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40
3
3
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40
add a comment |
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3
$begingroup$
Is this a maths question, or is this a Desmos question?
$endgroup$
– Lord Shark the Unknown
Dec 23 '18 at 5:21
$begingroup$
(1) is true if $xneq 1$ (you can simplify IF $xneq 1$)
$endgroup$
– Martín Vacas Vignolo
Dec 23 '18 at 5:35
$begingroup$
Left and right limits at $x=1$ exist, though the function is not defined at this point. So, consider the graph to have a hole at $(1,-1/4)$ (though the single point is not visible on desmos graph).
$endgroup$
– farruhota
Dec 23 '18 at 5:40