Find the values in the interval (1, 2) of the mean value theorem satis

1.	Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x2 for 1 ≤ x ≤ 2.
Answer» f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that f'(c) = \(\frac {f(2)−f(1)}{2 -1}\) = 1 – 2c That is, 1 – 2c = -2 ⇒ c = 3/2

Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x2 for 1 ≤ x ≤ 2.

Answer»

f(1) = 0 and f(2) = -2. Clearly f(x) is defined and differentiable in 1 < x < 2. Therefore, by the Mean Value Theorem, there exists a c ∈(1, 2) such that

f'(c) = \(\frac {f(2)−f(1)}{2 -1}\) = 1 – 2c

That is, 1 – 2c = -2

⇒ c = 3/2

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Find the values in the interval (1, 2) of the mean value theorem satisfied by the function f(x) = x – x2 for 1 ≤ x ≤ 2.

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