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hi are you there? if this is a y or f(x) then differentiate it then equaled it to zero and find its roots, that rot will be your max or min point..just plug it into your original equation....
yeah hehe. kay thanks (:
the derivative would be like y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,....NOW FIND ITS ROOTS OR ZEROES
TO FIND THE ROOTS HERE I THNK ITS EASIER TO GRAPH IT...OR USE SYNTHETIC DIVISION
doing synthetic division now...big numbers ):
lets try graphing the y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,...
can you tell what the asymptotic behavior with this equation is?
ok this is quartic equation,, so you have ti look for the family of curves of all the quartic equations behaviour....its like a sine wave but not quite as good as sine wave it wil touch the grap at x, 4 times in the graph
it will have 4 roots, and it will touch the x axis four times
can i post the graph of the equation?
did you graph it yet? the derivsative graph?
well it is a parent function picture so i took a picture with the shape of a quartic function and plotted points on the graph
ok get its roots and plug it in the original equation
are you familiar with newtons method of approximation?
if you have a calculus book its in there also its in the web
no whats that?
newtons method of aprox is good in finding he rots of equations..
oh well what do you have to do?
y'=-5.12x^3 -100.8x^2 -646.82X -1384.63,,....NOW FIND ITS ROOTS OR ZEROES ..to find its roots use newton method of approx.here you wil find 3 roots,then this 3 roots you will then plug them into your origimal equation to find you max y or min y
did you see the graph?
it didnot download in my laptop bec my c is full..here i can help u with newtons method,did you read the website i gave u?
I read the wikipedia one
newtons formula is x2=xi-[f(x1)/f'(x1)]
how do I get the actual coordinates?
here your f(x)=5.12x^3-100.8x^2-646.82x-1384.63 now use derivative to find f'(x)?
f'(x)=15.36x^2-201.6x-646.82 now you can use newtons method formula to find the roots of f(x)=5.12x^3-100.8x^2-646.82x-1384.63
newton formula is x2=x1-[f(x1)/f ' (x)] start with x1 like -8, -9, -7, hope you are using a good calculator with this calculation
using x2=x1- [(5.12x^2-100.8x^2-646.82x-1384.63)/(15.36x^2-201.6x-646.82)] i got a root of x=25.13527 now you can plug it on you original equation
x2=x1- [(5.12x^3-100.8x^2-646.82x-1384.63)/(15.36x^2-201.6x-646.82)] im sorry that was 5.12x^3
just plug the root x =25.13527 here, f(x)=-1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60
and that should get me my minimum/maximum?
is this your college subject calculus?
im in high school pre calculus!
i think in this problem they want you to learn how to graph this function, and to learn how to get the roots of the first derivative.....then to find its max or min y you will have to use the root to find the min n the max
i wish i could look at the graph you want to show me, is it the graph of y=f(x)=-1.28x^4 + -33.36x^3 + -323.41x^2 + -1384.63x + -2206.60..or the graph of y'=f '(x)=5.12x^3-100.8x^2-646.82x-1384.63 ..trace it at x=0, then you can find the y axis which is max or min
its the first one
yes only the first one bec x^4 is quartic therefore it must have 4 roots and 4 max or min y...correction on the first root i check it again on my calculator it is x=25.14043077
ook i have another question... how do i know which one is the positive/negative/complex roots?
so when you plug this x=25.14043077 to the orig eq f(25.043077)= -1282838.114
ypu notice that this is neg 1.million which is hard to graph,,therefore you will need to use the newton method formula..thats the conclusion here,,,you will be using the newton formula again and guess again for the roots
oh well thanks soo much for your help i really appreciate it!
hope you will follow the procedure again and again to find the other 3 roots....good luck ....is this in college subject? or ap calculus in high school?
umm i go to an option school, and we dont really have ap classes, but there is advanced math...thats what i take
oh ok....whews this is really for college problem hehehe thats really advance calculus
ook i have another question... how do i know which one is the positive/negative/complex roots? if you remember the quadratic formula x= [-b+-sqrt(b^2-4ac)/2a]...the equation is y=ax^2+bx+c
positive complex root is like x=(2+sqrt-4)/2 = (2+2i)/2= 1+i did you noticed the plus or positive sign? that a positive complex root..for conjugate or neg complex root its x=1-i