Question

Verify mean value theorem for the function f(x) = (x – 3)(x – 6)(x – 9) in [3, 5].

Answer 1

(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).