Question: The Second Derivative Test for critical points fails if (c) = 0. This exercise develops a Higher Derivative Test based on the sign of the

The Second Derivative Test for critical points fails if ƒ"(c) = 0. This exercise develops a Higher Derivative Test based on the sign of the first nonzero derivative. Suppose that

f'(c) = f'(c)= = f(n-) (c) = 0, but f(c) # 0

(a) Show, by applying L’Hôpital’s Rule n times, that

lim XC f(x) = f(c) (x-c)" 1 n! f(n) (c)

where n! = n(n − 1)(n − 2) · · · (2)(1).

(b) Use (a) to show that if n is even, then ƒ(c) is a local minimum if ƒ(n)(c) > 0 and is a local maximum if ƒ(n)(c)

(c) Use (a) to show that if n is odd, then ƒ(c) is neither a local minimum nor a local maximum.

f'(c) = f'(c)= = f(n-) (c) = 0, but f(c) # 0

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a b From part a lim yt lim C w w1 This may be rewritten as where cos At 1 1 C lim 61 ... View full answer

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