Question

determine the location of each local extrenum of the function f(X)= x^4/4 - 8X^3/3 +19X^2/2 -12X +2

Answer 1

take the derivative, set = 0, and solve. The derivative is a cubic polynomial, not easy to find the roots, but you can see by inspection that 1 is a solution. Factor (x-1) from your derivative so get
\[f'(x)=(x-1)(x^2-7x+12 )= (x-1)(x-4)(x-3))\]
therefore zeros are -1,3 and 4. -1 and 4 are relative min, as you can see if you draw a picture of a fourth degree polynomial with positive leading coefficient, and 3 is a relative max

Answer 2

oops my typo. zeros are 1, 3 and 4, not -1!

Answer 3

that's my problem i'm having trouble setting it to f' how did u get (x-1)(x^2-7x+12)??

Answer 4

that's my problem i'm having trouble setting it to f' how did u get (x-1)(x^2-7x+12)??

Answer 5

Find the derivative of each term. After that factor, like in college algebra or pre-cal.

Answer 6

ok

Answer 7

I got it by cheating. It is hard to find the zeros of a cubic polynomial but I figured that this problem was easy so I guessed 1. I checked that f'(1) = 0. Now "factor theorem" says if r is a zero of a polynomial p(x) then p(x) = (x-r)q(x) so I knew that f'(x) = (x-1) (something)
I found the "something" by synthetic division, but you can do long division as well.
http://www.youtube.com/watch?v=bZoMz1Cy1T4
Once you have a quadratic you can always find the zeros using the formula but this one factored easily.

Answer 8

ok thanx:)