3x^2-6x-3 = 0

Question

I'm tasked at finding the local Max. for a function f(x) = x^3 - 3x^2 - 3x -4

I'm looking for the critital points by finding where f'(x) = 0, or doesn't exhist.

f(x) = 3x^2 - 6x - 3

The solution incluces finding the "roots" and I don't understand that.   I think I'm forgeting basic algebra.    I need to know how to solve for X when 3x^2 - 6x - 3 = 0

Thanks in advance.

Britta

Answer 1

at best me kan ges bout the stuf yu rite...dy/dx=3x^2 -6x-3

& yu wanna set it tu zero

first chanj it tu 3*(x^2-2x-1)

then yuze quadratik equashun tu find zeroes...a=1, b=-2, c=-1

roots=2.414213562 & -0.414213562

b^2-41c=8, sqroot(8)=2*sqrt(2)=2.828427125

-b/2a=1, so root/2a=sqrt(2)=1.414213562373055048801688724205...