1. **State the problem:** Find the absolute minimum value of the function $f(x) = \cos(x)$ on the interval $[0, \pi]$. 2. **Recall the function and interval:** The cosine function $\cos(x)$ is continuous and differentiable on $[0, \pi]$. We want to find the lowest value it attains on this interval. 3. **Find critical points:** Take the derivative: $$f'(x) = -\sin(x)$$ Set $f'(x) = 0$ to find critical points: $$-\sin(x) = 0 \implies \sin(x) = 0$$ On $[0, \pi]$, $\sin(x) = 0$ at $x = 0$ and $x = \pi$. 4. **Evaluate $f(x)$ at critical points and endpoints:** - At $x=0$: $f(0) = \cos(0) = 1$ - At $x=\pi$: $f(\pi) = \cos(\pi) = -1$ 5. **Check values inside the interval:** Since the only critical points are at the endpoints, no other points to check. 6. **Determine absolute minimum:** Among $1$ and $-1$, the minimum value is $-1$ at $x=\pi$. **Final answer:** The absolute minimum value of $f(x) = \cos(x)$ on $[0, \pi]$ is **$-1$**. This corresponds to choice D) -1.