Math miscellaneousEasy
Question
Let f be differentiable for all x. If f(1) = - 2 and f′(x) ≥ 2 for x ∈ [1, 6], then
Options
A.f(6) ≥ 8
B.f(6) < 8
C.f(6) < 5
D.f(6) = 5
Solution
As f(1) = - 2 & f′(x) ≥ 2 ∀ x ∈ [1, 6]
Applying Lagrange′s mean value theorem
= f′(c) ≥ 2
⇒ f(6) ≥ 10 + f(1)
⇒ f(6) ≥ 10 - 2
⇒ f(6) ≥ 8
Applying Lagrange′s mean value theorem
= f′(c) ≥ 2⇒ f(6) ≥ 10 + f(1)
⇒ f(6) ≥ 10 - 2
⇒ f(6) ≥ 8
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