Verify that the hypotheses of Rolle’s Theorem are satisfied for f(x)=6cosx on the interval [9pi/2,11pi/2] and find all values of c in this interval that satisfy the conclusion of the theorem.
its differentiable at 9pi/2,11pi/2.derivertive of 6cosx its -6sinx,value of c=0.
its differentiable at 9pi/2,11pi/2.derivertive of 6cosx its -6sinx,
To verify that the hypotheses of Rolle's Theorem are satisfied for the function f(x) = 6cos(x) on the interval [9π/2, 11π/2], we need to check two conditions:
1. Continuity: f(x) must be continuous on the closed interval [9π/2, 11π/2].
2. Differentiability: f(x) must be differentiable on the open interval (9π/2, 11π/2).
Let's examine each condition individually:
1. Continuity:
First, note that the function f(x) = 6cos(x) is a trigonometric function, and all trigonometric functions are continuous for all real numbers. This means that f(x) = 6cos(x) is continuous on the entire real number line, including the interval [9π/2, 11π/2]. Therefore, the continuity condition is satisfied.
2. Differentiability:
To check differentiability, we need to take the derivative of f(x) and examine if it exists for all x ∈ (9π/2, 11π/2).
The derivative of f(x) = 6cos(x) is obtained by applying the chain rule. The derivative of cos(x) is -sin(x), so:
f'(x) = -6sin(x)
Now, we need to ensure that f'(x) exists for all x ∈ (9π/2, 11π/2). Since sin(x) is also a trigonometric function that is defined for all real numbers, the only potential problem is when sin(x) is undefined, which occurs when x = (2n+1)π/2 for any integer n.
In the interval (9π/2, 11π/2), sin(x) is always defined and continuous; therefore, f'(x) = -6sin(x) exists and is continuous within the interval.
Since both conditions of continuity and differentiability are satisfied, we can conclude that the hypotheses of Rolle's Theorem are met for f(x) = 6cos(x) on the interval [9π/2, 11π/2].
To find the values of c in the interval [9π/2, 11π/2] that satisfy the conclusion of Rolle's Theorem, we need to find the x-values where f'(c) = 0.
We have f'(x) = -6sin(x), so when -6sin(c) = 0, sin(c) must equal 0.
The values of x in the interval [9π/2, 11π/2] where sin(x) = 0 are x = 9π/2 and x = 11π/2, corresponding to the points c = 9π/2 and c = 11π/2.
Therefore, the conclusion of Rolle's Theorem is satisfied at the points c = 9π/2 and c = 11π/2.