Better method for graphing functions with greater complexity
$begingroup$
I spent perhaps 30 mins graphing the following function.
$$y = frac{1}{|{2x-x^2}|} + frac{1}{|{2x+x^2}|}$$
I proceeded by first graphing the first half of the function $y = frac{1}{|{2x-x^2}|}$.
Then I graphed the second part of the function $y = frac{1}{|{2x+x^2}|}$.
Then using both graphs I drew a combined version which I used as my answer to the exercise.
This process was fairly time consuming and I am wondering is there a more concise way of doing this? Is there a way of understanding the general shape of such functions like we understand the more basic functions such as $y = frac{1}{x}$ or $y = x^3$ and then just inputing some key input values to get a more precise description of the graph?
If I could have an intuition of more complex functions that are combinations of the more basic functions it would be much easier to do such graphing exercises. Is there a method for getting such intuition? Is there a more concise way to complete such graphing exercises?
graphing-functions
asked Jan 4 at 19:35
esc1234esc1234
83
$endgroup$
add a comment |
$begingroup$
I spent perhaps 30 mins graphing the following function.
$$y = frac{1}{|{2x-x^2}|} + frac{1}{|{2x+x^2}|}$$
I proceeded by first graphing the first half of the function $y = frac{1}{|{2x-x^2}|}$.
Then I graphed the second part of the function $y = frac{1}{|{2x+x^2}|}$.
Then using both graphs I drew a combined version which I used as my answer to the exercise.
This process was fairly time consuming and I am wondering is there a more concise way of doing this? Is there a way of understanding the general shape of such functions like we understand the more basic functions such as $y = frac{1}{x}$ or $y = x^3$ and then just inputing some key input values to get a more precise description of the graph?
If I could have an intuition of more complex functions that are combinations of the more basic functions it would be much easier to do such graphing exercises. Is there a method for getting such intuition? Is there a more concise way to complete such graphing exercises?
graphing-functions
asked Jan 4 at 19:35
esc1234esc1234
83
$endgroup$
$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05
add a comment |
$begingroup$
I spent perhaps 30 mins graphing the following function.
$$y = frac{1}{|{2x-x^2}|} + frac{1}{|{2x+x^2}|}$$
I proceeded by first graphing the first half of the function $y = frac{1}{|{2x-x^2}|}$.
Then I graphed the second part of the function $y = frac{1}{|{2x+x^2}|}$.
Then using both graphs I drew a combined version which I used as my answer to the exercise.
This process was fairly time consuming and I am wondering is there a more concise way of doing this? Is there a way of understanding the general shape of such functions like we understand the more basic functions such as $y = frac{1}{x}$ or $y = x^3$ and then just inputing some key input values to get a more precise description of the graph?
If I could have an intuition of more complex functions that are combinations of the more basic functions it would be much easier to do such graphing exercises. Is there a method for getting such intuition? Is there a more concise way to complete such graphing exercises?
graphing-functions
asked Jan 4 at 19:35
esc1234esc1234
83
$endgroup$
I spent perhaps 30 mins graphing the following function.
$$y = frac{1}{|{2x-x^2}|} + frac{1}{|{2x+x^2}|}$$
I proceeded by first graphing the first half of the function $y = frac{1}{|{2x-x^2}|}$.
Then I graphed the second part of the function $y = frac{1}{|{2x+x^2}|}$.
Then using both graphs I drew a combined version which I used as my answer to the exercise.
This process was fairly time consuming and I am wondering is there a more concise way of doing this? Is there a way of understanding the general shape of such functions like we understand the more basic functions such as $y = frac{1}{x}$ or $y = x^3$ and then just inputing some key input values to get a more precise description of the graph?
If I could have an intuition of more complex functions that are combinations of the more basic functions it would be much easier to do such graphing exercises. Is there a method for getting such intuition? Is there a more concise way to complete such graphing exercises?
graphing-functions
graphing-functions
asked Jan 4 at 19:35
esc1234esc1234
83
asked Jan 4 at 19:35
esc1234esc1234
83
asked Jan 4 at 19:35
esc1234esc1234
83
asked Jan 4 at 19:35
esc1234esc1234
83
asked Jan 4 at 19:35
esc1234esc1234
83
83
$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05
add a comment |
$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05
$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You need to consider four cases:
$x<-2$
$-2<x<0$;
$0<x<2$ and
$x>2$.
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
$endgroup$
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
add a comment |
$begingroup$
Abuse that:
$$|x|=x text { or } -x$$
Figure out when both expressions inside the moduli are positive or negative and graph those cases individually.
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
add a comment |
$begingroup$
Why not just plot it--using software available to all?
And let me strongly recommend anyone who thinks graphing "by hand" is in any way superior to graphing by computer (e.g., because it teaches "fundamentals" or "solidifies understanding"), see this important TED talk.
The OP spent "perhaps 30 min" trying to graph this function by hand and understand it. I spent less than one minute writing the below code. For those who can run Mathematica, try it and leave a comment as to whether or not you understand the underlying function better by interacting with the figure (for 30 minutes!) than by getting out paper and pencil.
Manipulate[
Plot[1/Abs[a x + x^2] + 1/Abs[b x - x^2],
{x, -5, 5}],
{{a, 2}, 1, 3}, {{b, 2}, 1, 3}]
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
1
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
1
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
|
show 4 more comments
$begingroup$
$2 x - x^2 = x (2 - x) = 0$ when $x=0$ or $2$, and similarly $2 x + x^2 = 0$ when $x = 0$ or $-2$.
$1/|2x - x^2|$ is always positive, goes to $0$ at $pm infty$ and to $+infty$ as $x to 0$ or $2$, so looks something like this:
(well, exactly like that, since I used software, but you could do a similar rough sketch by hand).
$1/|2x + x^2| = 1/|2(-x) - (-x)^2|$ is the reflection of this across the $y$ axis. It goes to $0$ at $pm infty$ and to $+infty$ as $x to -2$ or $0$.
Thus the sum of these looks something like
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to consider four cases:
$x<-2$
$-2<x<0$;
$0<x<2$ and
$x>2$.
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
$endgroup$
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
add a comment |
$begingroup$
You need to consider four cases:
$x<-2$
$-2<x<0$;
$0<x<2$ and
$x>2$.
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
$endgroup$
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
add a comment |
$begingroup$
You need to consider four cases:
$x<-2$
$-2<x<0$;
$0<x<2$ and
$x>2$.
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
$endgroup$
You need to consider four cases:
$x<-2$
$-2<x<0$;
$0<x<2$ and
$x>2$.
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
answered Jan 4 at 19:39
Michael RozenbergMichael Rozenberg
106k1894198
106k1894198
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
add a comment |
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
what your suggesting is in the event that the function to be graphed contains fractions, leverage that by finding out when the denominator is 0 and evaluate x-values within those intervals. thanks! But is there a general method or way of intuitively visualizing such complex graphs or is it a matter of evaluating x-values and grinding away?
$endgroup$
– esc1234
Jan 4 at 19:49
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
$begingroup$
@esc1234 With the absolute value this is the method, which you saw. In the general case I just don't know.
$endgroup$
– Michael Rozenberg
Jan 4 at 19:51
add a comment |
$begingroup$
Abuse that:
$$|x|=x text { or } -x$$
Figure out when both expressions inside the moduli are positive or negative and graph those cases individually.
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
add a comment |
$begingroup$
Abuse that:
$$|x|=x text { or } -x$$
Figure out when both expressions inside the moduli are positive or negative and graph those cases individually.
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
$endgroup$
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
add a comment |
$begingroup$
Abuse that:
$$|x|=x text { or } -x$$
Figure out when both expressions inside the moduli are positive or negative and graph those cases individually.
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
$endgroup$
Abuse that:
$$|x|=x text { or } -x$$
Figure out when both expressions inside the moduli are positive or negative and graph those cases individually.
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
answered Jan 4 at 19:41
Rhys HughesRhys Hughes
6,9571530
6,9571530
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
add a comment |
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
$begingroup$
yeah the absolute value just flips negative values around the x-axis. Im learning how this works through this graphing practice. The absolute value symbols make the function look more daunting but it's not that more difficult really
$endgroup$
– esc1234
Jan 4 at 19:53
add a comment |
$begingroup$
Why not just plot it--using software available to all?
And let me strongly recommend anyone who thinks graphing "by hand" is in any way superior to graphing by computer (e.g., because it teaches "fundamentals" or "solidifies understanding"), see this important TED talk.
The OP spent "perhaps 30 min" trying to graph this function by hand and understand it. I spent less than one minute writing the below code. For those who can run Mathematica, try it and leave a comment as to whether or not you understand the underlying function better by interacting with the figure (for 30 minutes!) than by getting out paper and pencil.
Manipulate[
Plot[1/Abs[a x + x^2] + 1/Abs[b x - x^2],
{x, -5, 5}],
{{a, 2}, 1, 3}, {{b, 2}, 1, 3}]
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
1
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
1
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
|
show 4 more comments
$begingroup$
Why not just plot it--using software available to all?
And let me strongly recommend anyone who thinks graphing "by hand" is in any way superior to graphing by computer (e.g., because it teaches "fundamentals" or "solidifies understanding"), see this important TED talk.
The OP spent "perhaps 30 min" trying to graph this function by hand and understand it. I spent less than one minute writing the below code. For those who can run Mathematica, try it and leave a comment as to whether or not you understand the underlying function better by interacting with the figure (for 30 minutes!) than by getting out paper and pencil.
Manipulate[
Plot[1/Abs[a x + x^2] + 1/Abs[b x - x^2],
{x, -5, 5}],
{{a, 2}, 1, 3}, {{b, 2}, 1, 3}]
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
1
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
1
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
|
show 4 more comments
$begingroup$
Why not just plot it--using software available to all?
And let me strongly recommend anyone who thinks graphing "by hand" is in any way superior to graphing by computer (e.g., because it teaches "fundamentals" or "solidifies understanding"), see this important TED talk.
The OP spent "perhaps 30 min" trying to graph this function by hand and understand it. I spent less than one minute writing the below code. For those who can run Mathematica, try it and leave a comment as to whether or not you understand the underlying function better by interacting with the figure (for 30 minutes!) than by getting out paper and pencil.
Manipulate[
Plot[1/Abs[a x + x^2] + 1/Abs[b x - x^2],
{x, -5, 5}],
{{a, 2}, 1, 3}, {{b, 2}, 1, 3}]
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
$endgroup$
Why not just plot it--using software available to all?
And let me strongly recommend anyone who thinks graphing "by hand" is in any way superior to graphing by computer (e.g., because it teaches "fundamentals" or "solidifies understanding"), see this important TED talk.
The OP spent "perhaps 30 min" trying to graph this function by hand and understand it. I spent less than one minute writing the below code. For those who can run Mathematica, try it and leave a comment as to whether or not you understand the underlying function better by interacting with the figure (for 30 minutes!) than by getting out paper and pencil.
Manipulate[
Plot[1/Abs[a x + x^2] + 1/Abs[b x - x^2],
{x, -5, 5}],
{{a, 2}, 1, 3}, {{b, 2}, 1, 3}]
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
edited Jan 5 at 0:23
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
answered Jan 4 at 19:41
David G. StorkDavid G. Stork
11k41432
11k41432
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
1
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
1
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
|
show 4 more comments
$begingroup$
If you're suggest graphing software, at least suggest the best one
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– Rhys Hughes
Jan 4 at 19:44
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Suggesting a software to graph a map on a math site is a funny joke!
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– mathcounterexamples.net
Jan 4 at 19:45
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the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
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– esc1234
Jan 4 at 19:45
1
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@RhysHughes: Um.... what did you mean by "abuse that"?
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– David G. Stork
Jan 4 at 19:46
1
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@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
If you're suggest graphing software, at least suggest the best one
$endgroup$
– Rhys Hughes
Jan 4 at 19:44
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
Suggesting a software to graph a map on a math site is a funny joke!
$endgroup$
– mathcounterexamples.net
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
$begingroup$
the exercise requires to do it by hand as a means of solidifying understanding. I could always cheat but I don't want to :)
$endgroup$
– esc1234
Jan 4 at 19:45
1
1
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
$begingroup$
@RhysHughes: Um.... what did you mean by "abuse that"?
$endgroup$
– David G. Stork
Jan 4 at 19:46
1
1
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
$begingroup$
@mathcounterexamples.net: The problem mentioned nothing whatsoever about plotting "by hand." 99.99% of math is done on computers now (when was the last time you inverted a matrix "by hand"!!?), and mathematicians working in most sub-disciplines who are not versed in tools such as Mathematica are the "funny joke". Are hand calculators "jokes" too?
$endgroup$
– David G. Stork
Jan 4 at 19:52
|
show 4 more comments
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$2 x - x^2 = x (2 - x) = 0$ when $x=0$ or $2$, and similarly $2 x + x^2 = 0$ when $x = 0$ or $-2$.
$1/|2x - x^2|$ is always positive, goes to $0$ at $pm infty$ and to $+infty$ as $x to 0$ or $2$, so looks something like this:
(well, exactly like that, since I used software, but you could do a similar rough sketch by hand).
$1/|2x + x^2| = 1/|2(-x) - (-x)^2|$ is the reflection of this across the $y$ axis. It goes to $0$ at $pm infty$ and to $+infty$ as $x to -2$ or $0$.
Thus the sum of these looks something like
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
add a comment |
$begingroup$
$2 x - x^2 = x (2 - x) = 0$ when $x=0$ or $2$, and similarly $2 x + x^2 = 0$ when $x = 0$ or $-2$.
$1/|2x - x^2|$ is always positive, goes to $0$ at $pm infty$ and to $+infty$ as $x to 0$ or $2$, so looks something like this:
(well, exactly like that, since I used software, but you could do a similar rough sketch by hand).
$1/|2x + x^2| = 1/|2(-x) - (-x)^2|$ is the reflection of this across the $y$ axis. It goes to $0$ at $pm infty$ and to $+infty$ as $x to -2$ or $0$.
Thus the sum of these looks something like
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
$endgroup$
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
add a comment |
$begingroup$
$2 x - x^2 = x (2 - x) = 0$ when $x=0$ or $2$, and similarly $2 x + x^2 = 0$ when $x = 0$ or $-2$.
$1/|2x - x^2|$ is always positive, goes to $0$ at $pm infty$ and to $+infty$ as $x to 0$ or $2$, so looks something like this:
(well, exactly like that, since I used software, but you could do a similar rough sketch by hand).
$1/|2x + x^2| = 1/|2(-x) - (-x)^2|$ is the reflection of this across the $y$ axis. It goes to $0$ at $pm infty$ and to $+infty$ as $x to -2$ or $0$.
Thus the sum of these looks something like
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
$endgroup$
$2 x - x^2 = x (2 - x) = 0$ when $x=0$ or $2$, and similarly $2 x + x^2 = 0$ when $x = 0$ or $-2$.
$1/|2x - x^2|$ is always positive, goes to $0$ at $pm infty$ and to $+infty$ as $x to 0$ or $2$, so looks something like this:
(well, exactly like that, since I used software, but you could do a similar rough sketch by hand).
$1/|2x + x^2| = 1/|2(-x) - (-x)^2|$ is the reflection of this across the $y$ axis. It goes to $0$ at $pm infty$ and to $+infty$ as $x to -2$ or $0$.
Thus the sum of these looks something like
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
answered Jan 4 at 20:25
Robert IsraelRobert Israel
326k23215469
326k23215469
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
add a comment |
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
$begingroup$
thanks for the explanation, the method you illustrated seems a good bit more efficient than the method I used described above.
$endgroup$
– esc1234
Jan 5 at 7:38
add a comment |
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$begingroup$
What were you ask precisely? Because taking 30 minutes to find the various cases, the local minimums, the inflection points if any, the asymptotes, the symmetries... is not so bad!
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– mathcounterexamples.net
Jan 4 at 19:49
$begingroup$
I was asked to "graph the following functions and specify whether each is odd or even"
$endgroup$
– esc1234
Jan 4 at 20:05