To evaluate f_{1}(x) = 2\sin(x) + 3\cos(x) for x = 0, \frac{\pi}{4}, \frac{\pi}{2}, we substitute these values into the function.
1. For x = 0,
f_{1}(0) = 2\sin(0) + 3\cos(0) = 2(0) + 3(1) = 3
2. For x = \frac{\pi}{4},
f_{1}\left(\frac{\pi}{4}\right) = 2\sin\left(\frac{\pi}{4}\right) + 3\cos\left(\frac{\pi}{4}\right) = 2\left(\frac{\sqrt{2}}{2}\right) + 3\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2} + \frac{3\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}
3. For x = \frac{\pi}{2},
f_{1}\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) + 3\cos\left(\frac{\pi}{2}\right) = 2(1) + 3(0) = 2
\textbf{Answer:}
f_{1}(0) = 3, \quad f_{1}\left(\frac{\pi}{4}\right) = \frac{5\sqrt{2}}{2}, \quad f_{1}\left(\frac{\pi}{2}\right) = 2