`f(x) = (x - 1)(x - 2)(x - 3), [1,3]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in...

`f(x) = (x - 1)(x - 2)(x - 3), [1,3]` Determine whether Rolle’s Theorem can be applied to `f` on the closed interval `[a, b]`. If Rolle’s Theorem can be applied, find all values of `c` in the open interval

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Chapter 3, 3.2 - Problem 11 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Yes it can.

The function f is continuous on [1, 3] and differentiable on (1, 3) (as an elementary function, or even as polynomial) and f(1) = f(3) (=0). All conditions are met.

The value of c is such that f'(c)=0.

f'(x) = (x-2)(x-3) + (x-1)*[(x-2)(x-3)]' =

(x-2)(x-3) + (x-1)*[(x-2) + (x-3)] =

x^2 - 5x + 6 + (x-1)*(2x-5) = x^2 - 5x + 6 + 2x^2 - 7x + 5 = 

3x^2 - 12x + 11.

This =0 for x_1,2 = [6 +- sqrt(36 - 33)]/3 = 2 +- sqrt(3)/3.
Both + and - options fit (1, 3).
(I used the formula for the even second factor.)

The answer: c_1 = 2 - sqrt(3)/3, c_2 = 2 + sqrt(3)/3.

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