solving inequalities (1 Viewer)

[...]

thats an awesome graph u have there...lol

 

[redslert]

wow if your impressed with that you should see the ones i draw in my exam

 

[SoCal]

Thanks heaps redslert, I can follow it all OK except how do you know what the graph looks like? You know the points that it cuts the x-axis but how do you know it does not go the other way around?

 

[ND]

Originally posted by redslert
you draw the graph here:

and then u would shade in everything up top because it is 'greater than' (>)

and then the answer will be


-1 <= x <= 0 & x >= 1


and the condition is that x cannot equal 1 or -1

.'. -1 < x <= 0 & x > 1

Nope that's not what the graph looks like. The graph f(x)=x/(1-x^2) has vertical asymptotes at x=+-1, as x -> 1, f(x) -> infinity, as x -> -1, f(x) -> -infinity... meh screw this explaining, let me get paint.

edit: nah i screwed my graph...

 

[...]

check

 

[ND]

No i don't think that's the right graph.

 

[ND]

Heres what i think it looks like (but i haven't done much graphing for a while so im not 100%).

 

[...]

ah, shit....

i have as equal..i might be the difference

 

[...]

actually, wat are we sketching...


???

 

[redslert]

hmmmmm well i dont know 'exactly' what the question was??

all i did was try to answer Merethrond as he wants the final solution to [ x(1-x)(1+x)>=0 ]

and that's the final as i assume the question was x / (1 - x^2) >= 0
and x cannot be -1 or +1


part 2
how did i get the graph????
well i dont know exactly how, i remember asking the exact same question a long time ago when i was learning this at the end of yr10
me sister told me to just memorise the graph like that if it is (+x) and if it a (-x) draw the graph the other way.....

there must be a logical explaination....
come to think of it....
it's like a graph of x^3 sideways... positive you turn clockwise
if negative you turn anti-clockwise.....

hmmm a bit confusing, i know...maths is difficult to explain....my approach is that everything is the way it is and don't question it....

turn to draw out x^3 and turn it clockwise/anti...you will get what i was tryin to say

 

[Dumbarse]

graph?

 

[ND]

Haha i think we're all sketching different graphs. Mine was f(x)=x/(1-x^2).

 

[Dumbarse]

same, but i cant use computers, can u see that graph?
i didnt know ho to paste a file or graph onto this so u can see it
cough retard cough

 

[...]

Originally posted by ND
Haha i think we're all sketching different graphs. Mine was f(x)=x/(1-x^2).

oh! x on top as well..i had 1 instead

let me re-do it

edit: yea, ND and Dumbarse looks right...

so whats the fuss??

 

[ND]

Originally posted by redslert
hmmmmm well i dont know 'exactly' what the question was??

all i did was try to answer Merethrond as he wants the final solution to [ x(1-x)(1+x)>=0 ]

and that's the final as i assume the question was x / (1 - x^2) >= 0
and x cannot be -1 or +1


part 2
how did i get the graph????
well i dont know exactly how, i remember asking the exact same question a long time ago when i was learning this at the end of yr10
me sister told me to just memorise the graph like that if it is (+x) and if it a (-x) draw the graph the other way.....

there must be a logical explaination....
come to think of it....
it's like a graph of x^3 sideways... positive you turn clockwise
if negative you turn anti-clockwise.....

hmmm a bit confusing, i know...maths is difficult to explain....my approach is that everything is the way it is and don't question it....

turn to draw out x^3 and turn it clockwise/anti...you will get what i was tryin to say

Now i see what you did. If you want to graph it, take it from the start. i.e. make it f(x)=x(1-x^2), then you can play with it all you want, and it'll still be the same graph. But what you did, was multiple through before making it f(x), which results in a completely different graph.

 

[ND]

Originally posted by Dumbarse
same, but i cant use computers, can u see that graph?
i didnt know ho to paste a file or graph onto this so u can see it
cough retard cough

Yeh mine is the same as yours.

 

[redslert]

why doesn't my solution work!!!!!!!
it's the other way around????
arrr

if i do my graph thing

the solution is: -1 < x <= 0 & x > 1

but it's the other way around with i put it to grafeq?!

hmmmmm :mad1:

 

[SoCal]

The original question was x/(1-x^2) >= 0 for all those who are still confused.

OK I ended up calling the HSC advice line and the answer is definately

x < -1 or 0 =< x < 1

The graph also looks like this:

 

[freaking_out]

Originally posted by Merethrond
The original question was x/(1-x^2) >= 0 for all those who are still confused.

OK I ended up calling the HSC advice line and the answer is definately

x < -1 or 0 =< x < 1

The graph also looks like this:

nice graph u've drawn there mate.

 

[SoCal]

Originally posted by freaking_out
nice graph u've drawn there mate.

Ha ha, I'm just trying to get the point across.