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Question
Verify mean value theorem for the function f(x) = (x – 3)(x – 6)(x – 9) in [3, 5].
Solution
(i) Function f is continuous in [3, 5] as product of polynomial functions is a polynomial, which is continuous.
(ii) f′(x) = 3x2 – 36x + 99 exists in (3, 5) and hence derivable in (3, 5).
Thus conditions of mean value theorem are satisfied.
Hence, there exists at least one c ∈ (3, 5) such that
f'(c) = `("f"(5) - "f"(3))/(5 - 3)`
⇒ 3c2 – 36c + 99 = `(8 - 0)/2` = 4
⇒ c = `6 +- sqrt(13/3)`.
Hence c = `6 +- sqrt(13/3)` .....(Since other value is not permissible).
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