Hi all, I am stuck at the end of this. Find the minimum value of the curve y = x⁴ - 2x² + 3 in the domain -3 ≤ x ≤ 3. y = x⁴ - 2x² + 3 y’ = 4x³ - 4x stationary points where y’ = 0 4x³ - 4x = 0 4(x³ - x) = 0 Stuck here
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4(x^3 - x) = 0 4x(x^2 - 1) = 0 Can you continue? ^^
Originally Posted by Joel Hi all, I am stuck at the end of this. Find the minimum value of the curve y = x⁴ - 2x² + 3 in the domain -3 ≤ x ≤ 3. y = x⁴ - 2x² + 3 y’ = 4x³ - 4x stationary points where y’ = 0 4x³ - 4x = 0 4(x³ - x) = 0 Stuck here Let . Then . You can find the minimum value of that by completing the square: The mimimum value of is 2 at u= 1 so the minimum value of is __________.
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