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Answers – Using Critical Points

1)
y’ = 6x + 4 = 0    x = -2/3 (minimum)
y” = 6 (no inflection points)
2)
y’ = 3x2 – 1/x = (3x3 – 1)/x = 0    x = (1/3)1/3 (minimum)
y” = 6x + 1/x2 = (6x3 + 1)/x2 = 0    x = (1/6)1/3 (inflection)
3)
y’ = 1 – cos x = 0    cos x = 1    x = 0 (neither – y” is also 0)
y” = -sin x = 0    x = 0 (inflection)
4)
y’ = 5x4 – 20x3 + 15x2 = 5x2(x2 – 4x + 3) = 0    x = 0 (neither), 1 (max), 3 (min)
y” = 20x3 – 60x2 + 30x = 20x(x2 – 3x + 1.5)    x = 0 (inflection), (1.5)1/2 (inflection) 

 

Copyright 2001 Bruce R.
Email: qrhetoric@yahoo.com