verify that the function satisfies the hypotheses of the mean value theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle’s Theorem.
f(x)=√x-1/3 x,[0,9]
(f(9)-f(0))/9 = (0-0)/9 = 0
Since f(0) = f(9) we can apply Rolle's Theorem.
f'(x) = 1/(2√x) - 1/3
So, look for some value x=c such that f'(c) = 0
1/(2√c) = 1/3
2√c = 3
c = 9/4
To verify that the function satisfies the hypotheses of the Mean Value Theorem (MVT), we need to check two conditions:
1. Continuity: The function f(x) must be continuous on the closed interval [0, 9].
2. Differentiability: The function f(x) must be differentiable on the open interval (0, 9).
Let's check these conditions:
1. Continuity:
The function f(x) = √(x-1)/3x is a composition of continuous functions (√x, x-1, 3x, and 1/3x), and since composition preserves continuity, f(x) is continuous on its domain.
2. Differentiability:
To check differentiability, let's compute the derivative of f(x).
f'(x) = [d/dx(√x)] * (1/3x) + (√x) * [d/dx(1/3x)]
= (1/2√x) * (1/3x) + (√x) * (-1/3x^2)
= 1/6√x + (-1/3√x)
= (-1/6√x).
The derivative f'(x) exists for all x ∈ (0, 9), except for x = 0. Thus, f(x) is differentiable on the open interval (0, 9).
Now, to find all numbers c that satisfy the conclusion of Rolle's Theorem, we need to check if there exists at least one point c in the open interval (0, 9) where f(c) = 0 and f'(c) = 0.
Since f(c) = √(c-1)/3c, to find the values of c where f(c) = 0, we solve the equation √(c-1)/3c = 0.
Setting the numerator equal to 0, we have c - 1 = 0, which gives c = 1. So, f(c) = 0 when c = 1.
To find the values of c where f'(c) = 0, we solve the equation (-1/6√c) = 0.
The only solution to this equation is c = 0. However, c = 0 is not in the open interval (0, 9), so it is not a valid solution.
Therefore, the only number c that satisfies the conclusion of Rolle's Theorem is c = 1.