Final answer:
To find the value of C using the Mean Value Theorem for the function f(x) on the interval [1,3], we first differentiate the function and then equate its derivative to the average rate of change over the interval. Solving for C gives us approximately 1.6505.
Step-by-step explanation:
The question asks to find the value of C using the Mean Value Theorem applied to the function f(x) = 9/x^3 on the interval [1,3]. The Mean Value Theorem states that if a function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one number c in the interval (a, b) such that f'(c) = [f(b) - f(a)] / [b - a]. To apply this here:
- First compute f'(x), the derivative of f(x).
- Next, find the difference quotient f(3) - f(1) over the interval [1, 3].
- Then set f'(c) equal to the difference quotient and solve for c.
f'(x) = -27/x^4. The difference quotient is f(3) - f(1) / 3 - 1, which simplifies to 9/27 - 9, or -8/3. Setting the derivative equal to this gives us -27/c^4 = -8/3. Solving for c yields c = root(4)27/2 which is approximately 1.6505. Therefore, the value of C is approximately 1.6505 on the interval [1, 3].