solving inequalities (1 Viewer)

[ND]

Originally posted by Merethrond


The graph also looks like this:

Believe me, the graph of f(x)=x/(1-x^2) does not look like that. That's a cubic curve.

 

[Dumbarse]

look at my graph above the x axis, it has that answer

 

[freaking_out]

Originally posted by Rahul
my approach....

1/x >= 1/(x+1)

multiply each side by x^2 and (x+1)^2

x.(x+1)^2 >= x^2.(x+1)
...
x(x+1) >= 0

then solve by graphing.
you can treat these as two different finction to make it easier to understand.
y=x(x+1) and g=0

now graph where 'y is greater than or equal to g'.

answer: x<=-1, x>=0

hey, rahul, u would have lost a mark there, remember, that x does not equal to 0 or -1!!!

so therefore (providing that ur working is correct) i think the answer will be: x<-1, x>0

 

[SoCal]

ND, the question is not f(x)=x/(1-x^2) it is x/(1-x^2) >= 0, would that make a difference to the way the graph looks because it is an inequality and not a equation that is = to something but grater than zero? If you can follow what I am saying.

Dumbarse, true your graph does have the correct answer, now I don't know why graph is correct? HSC Advice Line or the graph that looks right and everyone agrees on? Also, why don't the values below the x-axis count?

 

[ND]

Originally posted by Merethrond
ND, the question is not f(x)=x/(1-x^2) it is x/(1-x^2) >= 0, would that make a difference to the way the graph looks because it is an inequality and not a equation that is = to something but grater than zero? If you can follow what I am saying.

Dumbarse, true your graph does have the correct answer, now I don't know why graph is correct? HSC Advice Line or the graph that looks right and everyone agrees on? Also, why don't the values below the x-axis count?

There is no graph of x/(1-x^2) >= 0, just solutions to it. What we are graphing is f(x)=x/(1-x^2), and because we want the values greater than zero, we can obersve the parts of the graph above the x-axis (i.e. when f(x)>0).

According to your graph, when x=1 y=0 right? Well put x=1 into the function f(x)=x/(1-x^2), it is undefined for that value, hence the asymptote on the graph. I don't know what the person at the HSC advice line was doing, but look at either Dumbarses or my graph, they are correct.

 

[Rahul]

Originally posted by freaking_out
hey, rahul, u would have lost a mark there, remember, that x does not equal to 0 or -1!!!

so therefore (providing that ur working is correct) i think the answer will be: x<-1, x>0

yes my mistake...

gotta do more maths

 

[redslert]

hmmmmmmmmm
someone just disproved everything ive learn.....

just great

 

[redslert]

oh fuck me dead i made the fuckin simplest mistake!!!!!!!!!!

god im gonna kill myself now

see you guy later....................

 

[freaking_out]

Originally posted by redslert
hmmmmmmmmm
someone just disproved everything ive learn.....

just great

lol, lucky that didn't happen the night bfore the hsc.

oh fuck me dead i made the fuckin simplest mistake!!!!!!!!!!

god im gonna kill myself now

see you guy later....................

ok, cya...good luck!

 

[SoCal]

Originally posted by ND
There is no graph of x/(1-x^2) >= 0, just solutions to it. What we are graphing is f(x)=x/(1-x^2), and because we want the values greater than zero, we can obersve the parts of the graph above the x-axis (i.e. when f(x)>0).

According to your graph, when x=1 y=0 right? Well put x=1 into the function f(x)=x/(1-x^2), it is undefined for that value, hence the asymptote on the graph. I don't know what the person at the HSC advice line was doing, but look at either Dumbarses or my graph, they are correct.

OK, I understand now, I didn't understand what the guy was talking about anyway.

That is a really hard question though don't you think? There is no way I could do that in an exam.

 

[iambored]

Originally posted by Rahul
iambored, dont do it! i doubt you will get marks for the working. marks are often assiciated with some trick with inequalities(multiply by the square of the denominator).

but i was taught this way at school, and have seen it in some books. i've got marks for working at school if that's any consolation?? but thanks

ok i might be a bit late but i'll post my way now

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:

alright,

x does not equal +/- 1 (from the demoninator)

let x/(1-x^2) = 0
by multiplying denominator by 0 you are left with
x=0

draw number line:
___-1____0____1____
test points by subbing them into the original

test: -2
2/3 >= 0 yes
test:-1/2
-2/3 >= 0 no
test:1/2
2/3 >=0 yes
test:2
-2/3 >= 0 no

now test the actual points to see if it's equal to the other side
-1 will be undefined, as will1
0 = 0

therefore the answer is x<-1 and 0<=x<1

 

[iambored]

hey i didn't realise u guys were still posting on this... now i gotta go back and read another page

 

[freaking_out]

Originally posted by Merethrond
OK, I understand now, I didn't understand what the guy was talking about anyway.

That is a really hard question though don't you think? There is no way I could do that in an exam.

yeah, there are really hard question in the 50 tips book, as well as cambridge yr. 11 book.

 

[SoCal]

Originally posted by freaking_out
yeah, there are really hard question in the 50 tips book, as well as cambridge yr. 11 book.

Is that question only a Year 11 question!

 

[redslert]

Originally posted by iambored
test: -2
2/3 >= 0 yes
test:-1/2
-2/3 >= 0 no
test:1/2
2/3 >=0 yes
test:2
-2/3 >= 0 no

now test the actual points to see if it's equal to the other side
-1 will be undefined, as will1
0 = 0

trail and error

i not too keen on it...
not many people are

but if you like it then stick with it

 

[freaking_out]

Originally posted by Merethrond
Is that question only a Year 11 question!

yep, its part of prelim. course i think. (well its in the yr. 11 book anyway)

edit: 1300 posts!!

 

[iambored]

Originally posted by redslert
trail and error

i not too keen on it...
not many people are

but if you like it then stick with it

i was testing points on the line...



i might ring the advice line just to check, or some other books. i'm serious i was taught that way!

 

[freaking_out]

Originally posted by iambored
i was testing points on the line...

i might ring the advice line just to check, or some other books. i'm serious i was taught that way!

i was taught like u, but we always kept our inequality sign.

 

[iambored]

Originally posted by freaking_out
i was taught like u, but we always kept our inequality sign.

u were taught like me?? cool. yeah i kep the inequality sign as well. it's just that it's greater than or equal to, so i had to put both of them

 

[freaking_out]

Originally posted by iambored
u were taught like me?? cool. yeah i kep the inequality sign as well. it's just that it's greater than or equal to, so i had to put both of them

yeah, and sometimes u can also sketch the graph, to find out the "yes" bits.