Find the value of c which satisfies Rolle’s Theorem for the function f(x)=sin(x²) on (0,√π).

ok, if true, there must be two points that are the same, and somewhere in between those points, the derivaative must be zero.

sin(X^2)=0 At 0, and at sqrt(PI)
f'=2xcosx^2=0 at 0 and at sqrt(PI/2)
c=sqrt(PI/2)

To find the value of c that satisfies Rolle's theorem, we need to check two conditions:

1. The function f(x) must be continuous on the closed interval [0, √π].
2. The function f(x) must be differentiable on the open interval (0, √π).

Let's first check the continuity of f(x) on [0, √π]:

The function f(x) = sin(x²) is continuous on its domain since sin(x) is a continuous function and x² is also continuous. Therefore, f(x) = sin(x²) is continuous on the interval [0, √π].

Next, let's check the differentiability of f(x) on (0, √π):

To find the derivative of f(x), we can use the chain rule. Let's differentiate f(x) = sin(x²) with respect to x:

f'(x) = cos(x²) * 2x

Now, we need to find the critical points on the interval (0, √π) where the derivative equals zero:

0 = cos(c²) * 2c

To satisfy this equation, either cos(c²) = 0 or 2c = 0.

If cos(c²) = 0, it means c² is an odd multiple of π/2.

If 2c = 0, it means c = 0.

However, the interval (0, √π) does not contain any values where c² is an odd multiple of π/2.

Therefore, the only value of c that satisfies Rolle's theorem for the function f(x) = sin(x²) on the interval (0, √π) is c = 0.