1. We are asked to find the absolute minimum and maximum of the function $f(x) = x^2 + 4x + 5$ on the interval $-3 \leq x \leq 1$. 2. The formula to find absolute extrema on a closed interval is to check the critical points inside the interval and the endpoints. 3. First, find the derivative: $$f'(x) = 2x + 4$$ 4. Set the derivative equal to zero to find critical points: $$2x + 4 = 0 \implies x = -2$$ 5. Check if $x = -2$ is in the interval $[-3,1]$. It is. 6. Evaluate $f(x)$ at the critical point and endpoints: - $f(-3) = (-3)^2 + 4(-3) + 5 = 9 - 12 + 5 = 2$ - $f(-2) = (-2)^2 + 4(-2) + 5 = 4 - 8 + 5 = 1$ - $f(1) = (1)^2 + 4(1) + 5 = 1 + 4 + 5 = 10$ 7. The absolute minimum is $1$ at $x = -2$ and the absolute maximum is $10$ at $x = 1$. The answer for part a) is: absolute minimum $1$ at $x = -2$, absolute maximum $10$ at $x = 1$.