Find the value of c which satisfies Rolle's Theorem for the function
f(x)=sin(x^2)on (0, �ãpi).
f(0) = 0
f(√π) = 0
So, we need c in (0,π) such that f'(c) = 0
f'(x) = 2x cos(x^2)
2x cos(x^2) = 0
Clearly c=√(π/2) satisfies the theorem.
To find the value of c that satisfies Rolle's Theorem for the function f(x) = sin(x^2) on the interval (0, π), we need to check three conditions:
1. The function f(x) is continuous on the closed interval [0, π].
2. The function f(x) is differentiable on the open interval (0, π).
3. The function f(x) takes the same value at the endpoints of the interval.
Let's start by checking the first condition:
1. The function f(x) = sin(x^2) is continuous on the interval [0, π] because sine is a continuous function and x^2 is also continuous on the interval.
Next, we need to check the second condition:
2. To check the differentiability, we need to find the derivative of f(x). The derivative of sin(x^2) can be found by applying the chain rule.
Using the chain rule, we have:
f'(x) = cos(x^2) * 2x.
The derivative of f(x) is given by f'(x) = cos(x^2) * 2x, which is defined for all x in the open interval (0, π).
Lastly, we need to verify the third condition:
3. We need to check if f(0) = f(π).
Since f(x) = sin(x^2), we have:
f(0) = sin(0^2) = sin(0) = 0, and
f(π) = sin(π^2).
Since π is not an integer multiple of π (which would make sin(π) = 0), sin(π^2) is not zero.
Therefore, we can conclude that all three conditions are not satisfied by f(x) = sin(x^2) on the interval (0, π).
As a result, we cannot find a value of c that satisfies Rolle's Theorem for this function on the given interval.