Find the absolute maxima for the function: $f(x)=5 x+\sin (5 x)$ on the interval $[0,2 \pi]$.

Enter the exact answers.

Step 1 ：Find the derivative of the function \(f(x)=5x+\sin(5x)\). The derivative is \(f'(x)=5+5\cos(5x)\).

Step 2 ：Set the derivative equal to zero and solve for \(x\). This gives us \(5+5\cos(5x)=0\), which simplifies to \(\cos(5x)=-1\). Solving for \(x\) gives us \(x=\frac{\pi}{5}+ \frac{2\pi k}{5}\), where \(k\) is an integer.

Step 3 ：Find the values of \(x\) that fall within the interval \([0,2\pi]\). These are \(x=\frac{\pi}{5}, \frac{3\pi}{5}, \frac{5\pi}{5}, \frac{7\pi}{5}, \frac{9\pi}{5}\).

Step 4 ：Evaluate the function at the critical points and the endpoints of the interval. This gives us \(f(0)=0\), \(f(\frac{\pi}{5})=\pi\), \(f(\frac{3\pi}{5})=3\pi\), \(f(\frac{5\pi}{5})=5\pi\), \(f(\frac{7\pi}{5})=7\pi\), \(f(\frac{9\pi}{5})=9\pi\), and \(f(2\pi)=10\pi\).

Step 5 ：The maximum value of the function on the interval \([0,2\pi]\) is the largest of these values. The maximum value is \(\boxed{10\pi}\) at \(x=2\pi\).