1. **Problem statement:** Find the absolute minimum value of the function $$f(x) = x \sin x + \cos x$$ on the interval $$[0, \pi]$$. 2. **Formula and rules:** To find absolute extrema on a closed interval, evaluate the function at critical points (where $$f'(x) = 0$$) inside the interval and at the endpoints. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(x \sin x + \cos x) = \sin x + x \cos x - \sin x = x \cos x$$ 4. **Set derivative to zero to find critical points:** $$f'(x) = x \cos x = 0$$ This implies either $$x = 0$$ or $$\cos x = 0$$. 5. **Solve for $$x$$ in $$[0, \pi]$$:** - $$x = 0$$ - $$\cos x = 0 \Rightarrow x = \frac{\pi}{2}$$ 6. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x=0$$: $$f(0) = 0 \cdot \sin 0 + \cos 0 = 0 + 1 = 1$$ - At $$x=\frac{\pi}{2}$$: $$f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} \cdot \sin \frac{\pi}{2} + \cos \frac{\pi}{2} = \frac{\pi}{2} \cdot 1 + 0 = \frac{\pi}{2} \approx 1.5708$$ - At $$x=\pi$$: $$f(\pi) = \pi \cdot \sin \pi + \cos \pi = \pi \cdot 0 - 1 = -1$$ 7. **Compare values:** - $$f(0) = 1$$ - $$f\left(\frac{\pi}{2}\right) \approx 1.5708$$ - $$f(\pi) = -1$$ The absolute minimum value is $$-1$$ at $$x = \pi$$. **Final answer:** The function has an absolute minimum at $$x = \pi$$ (Option C).