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thats an awesome graph u have there...lol
solving inequalities (1 Viewer)
- Thread starter ratherbesleepin
- Start date
redslert
yes, my actual brain
wow if your impressed with that you should see the ones i draw in my exam 
SoCal
Hollywood
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?
? You know the points that it cuts the x-axis but how do you know it does not go the other way around?
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N
ND
Guest
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.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
edit: nah i screwed my graph...
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check
N
ND
Guest
No i don't think that's the right graph.
N
ND
Guest
Heres what i think it looks like (but i haven't done much graphing for a while so im not 100%).
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ah, shit....
i have as equal..i might be the difference
i have as equal..i might be the difference
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actually, wat are we sketching...
???
???
redslert
yes, my actual brain
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
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
Member
graph?
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N
ND
Guest
Haha i think we're all sketching different graphs. Mine was f(x)=x/(1-x^2).
Dumbarse
Member
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
i didnt know ho to paste a file or graph onto this so u can see it
cough retard cough
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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??
N
ND
Guest
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.
N
ND
Guest
Yeh mine is the same as yours.
redslert
yes, my actual brain
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:
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
Hollywood
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:
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
Saddam's new life
nice graph u've drawn there mate.
SoCal
Hollywood
Ha ha, I'm just trying to get the point acrossOriginally posted by freaking_out
nice graph u've drawn there mate.
.