## anonymous 5 years ago what is the maximums and minimums of the equation y=2.657x^3 + -1.48x^2 + .2819x + 7.7E-14?

1. anonymous

thats kinda complicated to figure out you should go to http://www.wolframalpha.com

2. anonymous

3. 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....

4. anonymous

Y'=7.7971X^2 -2.96X +.2819.....NOW THIS IS A QUADRATIC EQUATION,.YOU CAN SOLVE FOR THE ROOTS NOW

5. anonymous

this is supposed to be a quartic...

6. anonymous

no,,, quartic eq starts at say x^4,....cubic is x^3.... quadratic is x^2

7. anonymous

what do i do after i derive the equation to find the minimums and maximums?

8. anonymous

this does start at x^4...... -1.28x4 + -33.36x3 + -323.41x2 + -1384.63x + -2206.60

9. anonymous

no it doesn't

10. anonymous

it's 2.657x^3.....

11. anonymous

hehe look up at your prob here atart at y=x^3 it a cubic eq rt?

12. anonymous

wait who are you talking to?

13. anonymous

after you done the derivative, the derivative is now a quadratic y'= ...x^2

14. anonymous

then how do you find the minimums and maximums

15. anonymous

ok ..you equate the derivative to zero and find its roots

16. anonymous

that roots will become you max or min of your original equation

17. anonymous

what is the answer after equating the derivative

18. anonymous

ok, to find the roots, do you know the quadratic formila?

19. anonymous

x=(-b+$\sqrt{?}$b2-4ac)2a

20. anonymous

im sorry my pc equation writer does not write well here you a=7.797.....b=-2.96...c= .2819

21. anonymous

x=(2.96+-$\sqrt{?}$-.0303)/15.594

22. anonymous

did you get it? the roots are imaginaries...do you know what is imaginary number?

23. anonymous

obviously... a number that's imaginary like i for example or e

24. anonymous

im just asking what are the points (max and min) for my equation

25. anonymous

no...any square root of a negative number are imaginary number like square root of -1, sqrot of -2, squroot of - numbers

26. 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?

27. anonymous

Y'=7.7971X^2 -2.96X +.2819.....graph this and find its root or zeroes

28. 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=?

29. anonymous

min y=0.02137744667

30. anonymous

i graph it rt now and found the root x=0.8948 now you can plug it in to your orig prob.

31. anonymous