Let’s analyze the problem step by step.

Given:

1. $f(x)$: \begin{cases} 0 & \text{if } x < -3, \\ 3 & \text{if } -3 \leq x < -1, \\ -4 & \text{if } -1 \leq x < 4, \\ 0 & \text{if } x \geq 4. \end{cases}$$
2. $g(x) = \int_{-3}^{x} f(t) \, dt$.

We need to calculate the following:

• (a) $g(-4)$
• (b) $g(-2)$
• (c) $g(0)$
• (d) $g(5)$
• (e) The absolute maximum of $g(x)$.

Step 1: Analyze the function $f(x)$

• On the interval $(-\infty, -3)$: $f(x) = 0$.
• On the interval $[-3, -1)$: $f(x) = 3$.
• On the interval $[-1, 4)$: $f(x) = -4$.
• On the interval $[4, \infty)$: $f(x) = 0$.

Step 2: Compute $g(x)$ for each case

(a) $g(-4)$:

Since $x = -4 < -3$, $f(t) = 0$ over the interval $[-3, -4]$. Thus: $g(-4) = \int_{-3}^{-4} f(t) \, dt = 0.$

(b) $g(-2)$:

Here $-3 \leq -2 < -1$, so $f(t) = 3$ over the interval $[-3, -2]$. Compute the integral: $g(-2) = \int_{-3}^{-2} 3 \, dt = 3 \cdot [t]_{-3}^{-2} = 3 \cdot (-2 - (-3)) = 3 \cdot 1 = 3.$

(c) $g(0)$:

Here $0 \in [-1, 4)$, so split the integral into two parts:

1. Over $[-3, -1]$, $f(t) = 3$.
2. Over $[-1, 0]$, $f(t) = -4$.

First, compute the contribution from $[-3, -1]$: $\int_{-3}^{-1} 3 \, dt = 3 \cdot (-1 - (-3)) = 3 \cdot 2 = 6.$

Next, compute the contribution from $[-1, 0]$: $\int_{-1}^{0} -4 \, dt = -4 \cdot (0 - (-1)) = -4 \cdot 1 = -4.$

Adding these contributions: $g(0) = 6 + (-4) = 2.$

(d) $g(5)$:

Here $x = 5 \geq 4$, so split the integral into three parts:

1. Over $[-3, -1]$, $f(t) = 3$.
2. Over $[-1, 4]$, $f(t) = -4$.
3. Over $[4, 5]$, $f(t) = 0$.

The first two parts were already computed:

• From $[-3, -1]$: $\int_{-3}^{-1} 3 \, dt = 6$,
• From $[-1, 4]$: $\int_{-1}^{4} -4 \, dt = -4 \cdot (4 - (-1)) = -4 \cdot 5 = -20$.

Adding these: $g(5) = 6 + (-20) = -14.$

(e) Absolute maximum of $g(x)$:

The function $g(x)$ increases when $f(x) > 0$ (on $[-3, -1]$) and decreases when $f(x) < 0$ (on $[-1, 4]$). The maximum occurs at $x = -1$: $g(-1) = \int_{-3}^{-1} f(t) \, dt = \int_{-3}^{-1} 3 \, dt = 6.$

Thus, the absolute maximum of $g(x)$ is: $\text{Maximum value} = 6 \quad \text{at } x = -1.$

Final Answers:

(a) $g(-4) = 0$
(b) $g(-2) = 3$
(c) $g(0) = 2$
(d) $g(5) = -14$
(e) The absolute maximum occurs at $x = -1$ and is $6$.

Would you like a graph of $f(x)$ or further clarification on any step?

Here are 5 follow-up questions to deepen understanding:

1. How does the sign of $f(x)$ affect the behavior of $g(x)$?
2. What is the physical interpretation of $g(x)$ in terms of the area under the curve of $f(x)$?
3. How would $g(x)$ behave if $f(x)$ were shifted to the right?
4. How can you determine the critical points of $g(x)$ using $f(x)$?
5. What happens to $g(x)$ as $x \to -\infty$ or $x \to \infty$?

Tip: For piecewise-defined integrals, always split the integral according to intervals where the function is constant or changes behavior!