Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (frostbite):

A function is given by: Y=(x+3x^2+3x^3+10x^4)/(1+4x+x^2+4x^3+10x^4 ),x≥0 Explain and argue for that Y/x assume a global maximum for x>0 (have done that). And determine the maximum with 3 decimals (just setting a calculator to get a lot of digits or zoom very close to the graph is not acceptable).

OpenStudy (frostbite):

I was thinking about determine the decimals by using the Newton/Raphson-method

OpenStudy (anonymous):

did you try to graph this one? so that you can see the MAX so you said the Max is Y/X ?

OpenStudy (anonymous):

Y/x=(1+3x+3x^2+10x^3)/(1+4x+x^2+4x^3+10x^4 )

OpenStudy (anonymous):

there are local min and local max

OpenStudy (frostbite):

Yea beside we may not plot the function

OpenStudy (anonymous):

to find local max n min find the derivative and equate it to zero to find the x

OpenStudy (frostbite):

I add another comment that also is stated in the problem: "You must explain your decimals reliability"

OpenStudy (anonymous):

set dy/dx= (-100x^6-60x^5-92x^4+16x^3+27x^2+4x-1)/(10x^4+4x^3+x2+4x+1)^2=0 x=0.131 x=0.512

OpenStudy (anonymous):

try also the Newton/Raphson-method and compare

OpenStudy (frostbite):

I know that you very esay can find the values but just useing a caculator which i have done, but the whole idea is that we are going to estamate a aproximated number without useing electronic help

OpenStudy (frostbite):

by using*

OpenStudy (anonymous):

do ypu have a graphing calculator? can you use that?

OpenStudy (frostbite):

No, we not use a graph either.

OpenStudy (anonymous):

(just setting a calculator to get a lot of digits or zoom very close to the graph is not acceptable).--->>> isnt this means that graphic calcu is ok that you cant zoom it only....

OpenStudy (anonymous):

well if you cant use a grapher try the Newton/Raphson-method

OpenStudy (frostbite):

And that is what i can't not very good useing it.

OpenStudy (anonymous):

what the grapher can do for us is to help us find the answer and compare what you have done on your solutions..lol,,....well good luck now

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!