Let f ‘(x₀) = f ‘’ (x₀) = ... = f ^(n-1) (x₀) = 0, but f⁽ⁿ⁾ (x₀) ≠  0. Then

1.    If n is odd, f has neither a relative maximum nor a relative minimum at x = x₀.

2.    If n is even, f has a relative maximum if f ⁽ⁿ⁾ (x₀) < 0, and has a relative minimum if f ⁽ⁿ⁾ (x₀) > 0.

This method is illustrated in the following examples.

Example. Use the second derivative test to find the relative maxima and minima of f (x) = 2x³ – 15x² + 36x + 18.

Solution. Differentiating the given function with respect to x, we obtain

               f ‘ (x) = 6x² – 30x + 36 = 6 (x-2) (x-3)
Now        f ‘ (x) = 0 when x = 2 and x = 3
Also        f “ (x) = 12x – 30
               f “ (2) = 12 (2) – 30 = - 6 < 0

hence there is a relative maximum at x = 2.

Further    f “ (3) = 12 (3) – 30 = 6 > 0,
hence there is a relative minimum at x = 3.

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