Mixing
How many times will the difference of two functions intersect the summation of the same two functions?
$begingroup$
Given two functions $f(x)$ and $g(x)$,
$h(x) = f(x)-g(x)$
and
$i(x) = f(x) + g(x)$
can you know how many times $h(x)$ and $i(x)$ will intersect?
I ask this because I noticed that if $f(x)$ and $g(x)$ are linear, I only observed one point of intersection between $h(x)$ and $i(x)$.
When $f(x)$ and $g(x)$ are quadratic, I noticed two points of intersection between $h(x)$ and $i(x)$.
It seems like the number of intersections is related to the power of the functions $f(x)$ and $g(x)$ from which $h(x)$ and $i(x)$ are constructed from but I'm not sure how to prove it. Is there a way to do this?
functions polynomials
edited Jan 6 at 3:09
twnly
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
$endgroup$
add a comment |
$begingroup$
Given two functions $f(x)$ and $g(x)$,
$h(x) = f(x)-g(x)$
and
$i(x) = f(x) + g(x)$
can you know how many times $h(x)$ and $i(x)$ will intersect?
I ask this because I noticed that if $f(x)$ and $g(x)$ are linear, I only observed one point of intersection between $h(x)$ and $i(x)$.
When $f(x)$ and $g(x)$ are quadratic, I noticed two points of intersection between $h(x)$ and $i(x)$.
It seems like the number of intersections is related to the power of the functions $f(x)$ and $g(x)$ from which $h(x)$ and $i(x)$ are constructed from but I'm not sure how to prove it. Is there a way to do this?
functions polynomials
edited Jan 6 at 3:09
twnly
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
$endgroup$
$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
1
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23
add a comment |
$begingroup$
Given two functions $f(x)$ and $g(x)$,
$h(x) = f(x)-g(x)$
and
$i(x) = f(x) + g(x)$
can you know how many times $h(x)$ and $i(x)$ will intersect?
I ask this because I noticed that if $f(x)$ and $g(x)$ are linear, I only observed one point of intersection between $h(x)$ and $i(x)$.
When $f(x)$ and $g(x)$ are quadratic, I noticed two points of intersection between $h(x)$ and $i(x)$.
It seems like the number of intersections is related to the power of the functions $f(x)$ and $g(x)$ from which $h(x)$ and $i(x)$ are constructed from but I'm not sure how to prove it. Is there a way to do this?
functions polynomials
edited Jan 6 at 3:09
twnly
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
$endgroup$
Given two functions $f(x)$ and $g(x)$,
$h(x) = f(x)-g(x)$
and
$i(x) = f(x) + g(x)$
can you know how many times $h(x)$ and $i(x)$ will intersect?
I ask this because I noticed that if $f(x)$ and $g(x)$ are linear, I only observed one point of intersection between $h(x)$ and $i(x)$.
When $f(x)$ and $g(x)$ are quadratic, I noticed two points of intersection between $h(x)$ and $i(x)$.
It seems like the number of intersections is related to the power of the functions $f(x)$ and $g(x)$ from which $h(x)$ and $i(x)$ are constructed from but I'm not sure how to prove it. Is there a way to do this?
functions polynomials
functions polynomials
edited Jan 6 at 3:09
twnly
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
edited Jan 6 at 3:09
twnly
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
edited Jan 6 at 3:09
twnly
697112
edited Jan 6 at 3:09
twnly
697112
edited Jan 6 at 3:09
twnly
697112
697112
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
asked Jan 6 at 1:57
S.CramerS.Cramer
13618
13618
$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
1
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23
add a comment |
$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
1
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23
$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
1
1
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$f(x)+g(x)=f(x)-g(x) to 2g(x)=0 to g(x)=0$$
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
add a comment |
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$begingroup$
$$f(x)+g(x)=f(x)-g(x) to 2g(x)=0 to g(x)=0$$
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
add a comment |
$begingroup$
$$f(x)+g(x)=f(x)-g(x) to 2g(x)=0 to g(x)=0$$
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
add a comment |
$begingroup$
$$f(x)+g(x)=f(x)-g(x) to 2g(x)=0 to g(x)=0$$
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
$endgroup$
$$f(x)+g(x)=f(x)-g(x) to 2g(x)=0 to g(x)=0$$
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
answered Jan 6 at 2:22
Rhys HughesRhys Hughes
6,0171530
6,0171530
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
add a comment |
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
$begingroup$
Yes. Try it if you like.
$endgroup$
– Rhys Hughes
Jan 6 at 4:39
add a comment |
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$begingroup$
Functions don't intersect; it is their graphs that intersect. The graphs intersect when the functions are equal. $h$ and $i$ are equal precisely when $g=0$.
$endgroup$
– Gerry Myerson
Jan 6 at 2:04
$begingroup$
so if g(x) is a parabola that is vertically shifted above the x axis, then h(x) and i(x) will never intersect...regardless of what f(x) is?
$endgroup$
– S.Cramer
Jan 6 at 2:33
1
$begingroup$
The reason you see more points of intersection when the degree of $g(x)$ goes up is related to the Fundamental Theorem of Algebra: a polynomial of degree $k$ has $k$ (not necessarily real nor distinct) roots.
$endgroup$
– obscurans
Jan 6 at 3:10
$begingroup$
$h(x)$ and $i(x)$ will never intersect, no matter what $f(x)$ and $g(x)$ are. Functions don't intersect. Graphs of functions might intersect, but functions don't.
$endgroup$
– Gerry Myerson
Jan 6 at 3:23