Need help again : (
Anyone has any idea why is [5(x-2)] / [3(cuberootx)] < 0 for all x<0 ?
Thanks in advance as usual.
is the question : 5(x-2) / 3x^3 < 0x < 0
Originally posted by ƒlame:is the question : 5(x-2) / 3x^3 < 0x < 0
No. I am suppose to sketch the above out.
I have no idea why is it <0 for all x<0 though.
x<0 cannot cube root what unless using all the advance maths stuff which i hate.
Got real and imaginary.
Originally posted by dkcx:x<0 cannot cube root what unless using all the advance maths stuff which i hate.
I see this i really @___@
http://www.wolframalpha.com/input/?i=d/dx+[x^(5/3)-5(x)^(2/3)]
How did the graph turns out to be like this
I got test again next week, my "A"s depend on you already
Originally posted by deepak.c:
Got real and imaginary.
How to distinguish between real and imaginary? : D
I forget all about complex numbers liao
Tks in advance. : D
Originally posted by Forbiddensinner:I see this i really @___@
http://www.wolframalpha.com/input/?i=d/dx+[x^(5/3)-5(x)^(2/3)]
How did the graph turns out to be like this
I got test again next week, my "A"s depend on you already
Easy, u come i wack u till u cannot take the test then u will get an A..........bs for it :p
Originally posted by Forbiddensinner:How to distinguish between real and imaginary? : D
I forget all about complex numbers liao
Tks in advance. : D
I refunded back to teacher liao.
Originally posted by dkcx:Easy, u come i wack u till u cannot take the test then u will get an A..........bs for it :p
Later you wack ler i got A+ then how?
Then later my educational modules also A+ as well
Then you will get upset and wack me somemore, then my results will keep shooting up till I get 100%
Originally posted by Forbiddensinner:Later you wack ler i got A+ then how?
Then later my educational modules also A+ as well
Then you will get upset and wack me somemore, then my results will keep shooting up till I get 100%
ok set. Can A 2pm tml. Come, i get ready my equipment.
Originally posted by Forbiddensinner:Need help again : (
Anyone has any idea why is [5(x-2)] / [3(cuberootx)] < 0 for all x<0 ?
Thanks in advance as usual.
Hi,
When x < 0, x^(1/3) is negative.
When x < 0, x - 2 < 0 so 5(x - 2) is negative.
The result should then be positive.
Thanks!
Cheers,
Wen Shih
Originally posted by Forbiddensinner:Need help again : (
Anyone has any idea why is [5(x-2)] / [3(cuberootx)] < 0 for all x<0 ?
Thanks in advance as usual.
Hi Forbiddensinner,
Perhaps there is a problem with the question?
Cheers.