The graph of f is shown. Apply Rolle's Theorem and find all values of c such that f '(c) = 0 at some point between the labeled intercepts. (Enter your answers as a comma-separated list.)
f (x) = sin(2x)
To apply Rolle's Theorem, we need to check if the given function \(f(x) = \sin(2x)\) satisfies the conditions of the theorem.
Rolle's Theorem states that if a function is continuous on a closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one value \(c\) in the interval \((a, b)\) such that \(f'(c) = 0\).
Looking at the graph of \(f(x) = \sin(2x)\), we can see that between any two consecutive intercepts (where the function crosses the x-axis), the function passes through \(y = 0\) at least once. Moreover, since \(f(x) = \sin(2x)\) is a sinusoidal function with a period of \(\pi\), we can conclude that between any two consecutive intercepts, there are infinitely many points where \(f'(c) = 0\) for some value of \(c\).
Therefore, for the given function \(f(x) = \sin(2x)\), there are infinitely many values of \(c\) between the labeled intercepts where \(f'(c) = 0\). We can't list all those values as a comma-separated list since there are infinitely many of them.