Problem Overview:

We are given a graph of $f(x)$ on the interval $[-2, 2]$, and we define $g(x) = \int_{-2}^x f(t) \, dt$. The goal is to find where the absolute minimum of $g(x)$ occurs on $[-2, 2]$.

Step-by-step Solution:

1. Understand $g(x)$:

• $g(x)$ represents the accumulated area under the curve $f(t)$ from $-2$ to $x$.
• Positive areas (above the $x$-axis) increase $g(x)$, and negative areas (below the $x$-axis) decrease $g(x)$.

2. Analyze the graph of $f(x)$:

• From $x = -2$ to $x = -1$: $f(x)$ is negative and linear (slope = 1). The area here is a triangle below the $x$-axis with base 1 and height 1: $\text{Area} = -\frac{1}{2}(1)(1) = -\frac{1}{2}.$

• From $x = -1$ to $x = 1$: $f(x)$ is positive and linear (slope = 1). The area here is a triangle above the $x$-axis with base 2 and height 2: $\text{Area} = \frac{1}{2}(2)(2) = 2.$

• From $x = 1$ to $x = 2$: $f(x)$ is constant and equal to 1. The area here is a rectangle: $\text{Area} = (1)(1) = 1.$

3. Compute $g(x)$ at key points:

To identify the absolute minimum, compute $g(x)$ at key points where $x = -2, -1, 0, 1, 2$.

1. At $x = -2$: $g(-2) = \int_{-2}^{-2} f(t) \, dt = 0$.

2. At $x = -1$: $g(-1) = \int_{-2}^{-1} f(t) \, dt = -\frac{1}{2}$.

3. At $x = 0$: $g(0) = \int_{-2}^0 f(t) \, dt = -\frac{1}{2} + \frac{1}{2}(1)(1) = 0$.

4. At $x = 1$: $g(1) = \int_{-2}^1 f(t) \, dt = -\frac{1}{2} + 2 = \frac{3}{2}$.

5. At $x = 2$: $g(2) = \int_{-2}^2 f(t) \, dt = -\frac{1}{2} + 2 + 1 = \frac{5}{2}$.

4. Determine the absolute minimum:

• From the computed values, $g(x)$ has the following values: $g(-2) = 0, \quad g(-1) = -\frac{1}{2}, \quad g(0) = 0, \quad g(1) = \frac{3}{2}, \quad g(2) = \frac{5}{2}.$
• The absolute minimum occurs at $x = -1$, where $g(x) = -\frac{1}{2}$.

Final Answer:

The absolute minimum of $g(x)$ on $[-2, 2]$ occurs at $\boxed{x = -1}$.

Do you want further clarification or details?

Expanding the Concept:

1. What happens to $g(x)$ at other points on the interval $[-2, 2]$?
2. Why does the minimum of $g(x)$ correspond to the most negative accumulated area?
3. How would $g(x)$ behave if $f(x)$ were entirely negative on $[-2, 2]$?
4. What does the Fundamental Theorem of Calculus say about $g'(x)$ in this problem?
5. How could we solve this problem without explicitly calculating areas?

Tip:

Always relate the graph of $f(x)$ to changes in $g(x)$: positive $f(x)$ increases $g(x)$, while negative $f(x)$ decreases it.