f(x)=x^3+x^23x+4
Let the zeroes of the function above be a, b and c
Find f'(a)f'(b)f'(c)
From a graph there is one real zero
a = 2.677993482
I used synthetic division to reduce to a quadratic, then the quadratic formula gives complex conjugates b and c:
.8389967417 +/ .8886732139i
f'(x)=3x^2+2x3
f'(a)=13.159
f'(b)=1.579+6.25i
f'(c)=1.5796.25i
f'(a)f'(b)f'(c) = 547
Note: These answers are accurate to ten digits or so. I used my calculator's memory to hold them.

Posted by Jer
on 20161205 10:01:30 