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Vieta's derivatives (Posted on 2016-12-02) Difficulty: 4 of 5

f(x)=x^3+x^2-3x+4

Let the zeroes of the function above be a, b and c

Find f'(a)f'(b)f'(c)

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Calculator solution | Comment 1 of 2
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+2x-3
f'(a)=13.159
f'(b)=-1.579+6.25i
f'(c)=-1.579-6.25i

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 2016-12-05 10:01:30
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