anonymous
  • anonymous
what is the maximums and minimums of the equation y=2.657x^3 + -1.48x^2 + .2819x + 7.7E-14?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
thats kinda complicated to figure out you should go to http://www.wolframalpha.com
anonymous
  • anonymous
your welcome
anonymous
  • anonymous
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....

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anonymous
  • anonymous
Y'=7.7971X^2 -2.96X +.2819.....NOW THIS IS A QUADRATIC EQUATION,.YOU CAN SOLVE FOR THE ROOTS NOW
anonymous
  • anonymous
this is supposed to be a quartic...
anonymous
  • anonymous
no,,, quartic eq starts at say x^4,....cubic is x^3.... quadratic is x^2
anonymous
  • anonymous
what do i do after i derive the equation to find the minimums and maximums?
anonymous
  • anonymous
this does start at x^4...... -1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60
anonymous
  • anonymous
no it doesn't
anonymous
  • anonymous
it's 2.657x^3.....
anonymous
  • anonymous
hehe look up at your prob here atart at y=x^3 it a cubic eq rt?
anonymous
  • anonymous
wait who are you talking to?
anonymous
  • anonymous
after you done the derivative, the derivative is now a quadratic y'= ...x^2
anonymous
  • anonymous
then how do you find the minimums and maximums
anonymous
  • anonymous
ok ..you equate the derivative to zero and find its roots
anonymous
  • anonymous
that roots will become you max or min of your original equation
anonymous
  • anonymous
what is the answer after equating the derivative
anonymous
  • anonymous
ok, to find the roots, do you know the quadratic formila?
anonymous
  • anonymous
x=(-b+\[\sqrt{?}\]b2-4ac)2a
anonymous
  • anonymous
im sorry my pc equation writer does not write well here you a=7.797.....b=-2.96...c= .2819
anonymous
  • anonymous
x=(2.96+-\[\sqrt{?}\]-.0303)/15.594
anonymous
  • anonymous
did you get it? the roots are imaginaries...do you know what is imaginary number?
anonymous
  • anonymous
obviously... a number that's imaginary like i for example or e
anonymous
  • anonymous
im just asking what are the points (max and min) for my equation
anonymous
  • anonymous
no...any square root of a negative number are imaginary number like square root of -1, sqrot of -2, squroot of - numbers
anonymous
  • anonymous
ok here the simplest way of geting the derivative roots is to graph the derivative and approximate its roots..did you get what i mean?
anonymous
  • anonymous
Y'=7.7971X^2 -2.96X +.2819.....graph this and find its root or zeroes
anonymous
  • anonymous
here one root is 0.2819...you plug it in your orig equation y=2.657(.2819)^3 + -1.48(.2819)^2 + .2819(.2819)+ 7.7E-14=?
anonymous
  • anonymous
min y=0.02137744667
anonymous
  • anonymous
i graph it rt now and found the root x=0.8948 now you can plug it in to your orig prob.
anonymous
  • anonymous
your max or min y=0.9708296575
anonymous
  • anonymous
sorry never mind that root x=.2819 or min y=.02137744667....thats a mistake,, graph it and found that the max or min y=.9708296575

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