MonotonicityHard

Question

Consider the function for x = [-2, 3], f(x) =  if x ≠ 1, then

Options

A.f is discontinuous at x = 1 ⇒ Rolle′s theorem is not applicable in [-2, 3]
B.f(-2) ≠ f(3) ⇒ Rolle′s theorem is not applicable in [-2, 3]
C.f is not derivable in (-2, 3) ⇒ Rolle′s theorem is not applicable
D.Rolle′s theorem is applicable as f satisfies all the conditions and c of Rolle′s theorem is 1/2

Solution

f(-2) = f(3) = 0
f(x) is continuous in [-2, 3] & derivable in (-2,3) so Rolle′s theorem is applicable. so ∃ c ∈  (-2, 3) such that f′(c) = 0
⇒  = 0 ⇒  c = 1/2

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