f(x) = x3 - x2 - 6x + 6

[0, 3]

Just can't seem to get the right answer. Your help is greatly appreciated!! Thanks.

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f(x) = x3 - x2 - 6x + 6

[0, 3]

Just can't seem to get the right answer. Your help is greatly appreciated!! Thanks.

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## Answers (1)

f(x) = x³ - x² - 6x + 6

f(0) = 6

f(3) = 6

The function is continuous on [0, 3] and differentiable on (0, 3) and

f(0) = f(3). Therefore the function satisfies Rolle's Theorem on the interval [0, 3].

Find c ∈(0, 3) such that f'(c) = 0.

f'(c) = 3c² - 2c - 6 = 0

c = [2 ± √(2² + 4*3*6)] / 6 = [2 ± √(4 + 72)] / 6 = [2 ± √76] / 6

= [2 ± 2√19] / 6 = [1 ± √19] / 3

Only one value for c is in the interval (0, 3), namely:

c = [1 + √19] / 3