Question 36: Knowing the function y = x3 – 3x + 1 has two extreme points x1; x2 . Sum x12 + x22.
Determination set: CHEAP
y = x3 – 3x + 1 ⇒ y’ = 3×2 – 3 ⇔ x = ± 1.
So two extreme points are satisfied: xfirst2 + x22 = 2
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Question 36: Knowing the function y = x3 – 3x + 1 has two extreme points x1; x2 . Sum x12 + x22.
Determination set: CHEAP
y = x3 – 3x + 1 ⇒ y’ = 3×2 – 3 ⇔ x = ± 1.
So two extreme points are satisfied: xfirst2 + x22 = 2
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