Maths Help

Forbiddensinner SG

Need help again : (

Anyone has any idea why is [5(x-2)] / [3(cuberootx)] < 0 for all x<0 ?

Thanks in advance as usual.

ƒlame SG

is the question : 5(x-2) / 3x^3 < 0x < 0

Forbiddensinner SG

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.

dkcx SG

x<0 cannot cube root what unless using all the advance maths stuff which i hate.

deepak.c SG

Got real and imaginary.

Forbiddensinner SG

x<0 cannot cube root what unless using all the advance maths stuff which i hate.

I see this i really @[email protected]

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

Forbiddensinner SG

Got real and imaginary.

How to distinguish between real and imaginary? : D

I forget all about complex numbers liao

Tks in advance. : D

dkcx SG

I see this i really @[email protected]

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

deepak.c SG

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.

Forbiddensinner SG

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%

dkcx SG

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.

wee_ws SG

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

TrueHeart SG

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.