1. **State the problem:** Find the absolute minimum and absolute maximum of the function $$f(x) = x^2 + 4x + 5$$ on the interval $$[-3, 1]$$. 2. **Formula and rules:** To find absolute extrema on a closed interval, evaluate the function at critical points inside the interval and at the endpoints. 3. **Find critical points:** Compute the derivative: $$f'(x) = 2x + 4$$ Set derivative to zero to find critical points: $$2x + 4 = 0 \implies x = -2$$ 4. **Check if critical point is in the interval:** $$-3 \leq -2 \leq 1$$, so yes. 5. **Evaluate function at critical point and endpoints:** - At $$x = -3$$: $$f(-3) = (-3)^2 + 4(-3) + 5 = 9 - 12 + 5 = 2$$ - At $$x = -2$$: $$f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$$ - At $$x = 1$$: $$f(1) = 1^2 + 4(1) + 5 = 1 + 4 + 5 = 10$$ 6. **Determine absolute extrema:** - Absolute minimum is $$1$$ at $$x = -2$$. - Absolute maximum is $$10$$ at $$x = 1$$. **Final answer:** - Absolute minimum: $$f(-2) = 1$$ - Absolute maximum: $$f(1) = 10$$