If the function f(x) = ax3 – 6x2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3], then the value of a must be ________.
Question
If the function f(x) = ax3 – 6x2 + 11x – 6 satisfies conditions of Rolle’s theorem in [1, 3], then the value of a must be ________.
Solution
Sure, let's solve this step by step.
Rolle's Theorem states that if a function f(x) is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the interval (a, b) such that the derivative f'(c) = 0.
Given the function f(x) = ax^3 – 6x^2 + 11x – 6, we need to find the value of 'a' for which the function satisfies the conditions of Rolle’s theorem in the interval [1, 3].
Step 1: Check if f(x) is continuous and differentiable on the interval [1, 3]. Since f(x) is a polynomial function, it is continuous and differentiable for all real numbers, including on the interval [1, 3].
Step 2: Check if f(1) = f(3). Substituting x = 1 and x = 3 into the function, we get:
f(1) = a(1)^3 – 6(1)^2 + 11(1) – 6 = a - 6 + 11 - 6 = a - 1 f(3) = a(3)^3 – 6(3)^2 + 11(3) – 6 = 27a - 54 + 33 - 6 = 27a - 27
For the function to satisfy Rolle's theorem, f(1) must equal f(3), so we set these equal to each other and solve for 'a':
a - 1 = 27a - 27 26a = 26 a = 1
So, the value of 'a' that makes the function f(x) = ax^3 – 6x^2 + 11x – 6 satisfy the conditions of Rolle’s theorem in the interval [1, 3] is a = 1.
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