1. **Problem:** Find local and global extrema of $f(x) = 4 - x^2$ on the interval $[-3, 1]$. 2. **Formula and rules:** To find extrema, first find critical points by setting the derivative $f'(x)$ to zero or undefined, then evaluate $f(x)$ at critical points and endpoints. 3. **Find derivative:** $$f'(x) = -2x$$ 4. **Find critical points:** Set $f'(x) = 0$: $$-2x = 0 \implies x = 0$$ 5. **Evaluate $f(x)$ at critical points and endpoints:** $$f(-3) = 4 - (-3)^2 = 4 - 9 = -5$$ $$f(0) = 4 - 0^2 = 4$$ $$f(1) = 4 - 1^2 = 4 - 1 = 3$$ 6. **Determine extrema:** - Local maximum at $x=0$ with value $4$ (since $f'(x)$ changes from positive to negative). - Global maximum is also $4$ at $x=0$. - Global minimum is $-5$ at $x=-3$. Final answer: Local and global maximum at $(0,4)$, global minimum at $(-3,-5)$.