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We are given a graph of \( g'(x) \), the derivative of \( g(x) \), and we need to find \( g(3) \) and \( g(-2) \). Additionally, we know that \( g(0) = 5 \).

To solve this, we will use the **Fundamental Theorem of Calculus**, which states:

\[
g(x) = g(0) + \int_0^x g'(t) \, dt
\]

Thus, to find \( g(3) \) and \( g(-2) \), we will integrate \( g'(x) \) over the respective intervals. This will require calculating the areas under the curve of \( g'(x) \), with the convention that areas above the \( x \)-axis contribute positively and areas below the \( x \)-axis contribute negatively.

### Finding \( g(3) \):
1. From \( x = 0 \) to \( x = 3 \), we integrate the graph of \( g'(x) \).
2. The graph includes a semicircle and a triangle. 
    - The semicircle has a radius of 1 and is above the x-axis (centered at \( x = 0 \)).
    - The triangle between \( x = 1 \) and \( x = 3 \) is also above the x-axis.

#### Step-by-step:
1. **Semicircle (from \( x = 0 \) to \( x = 1 \)):**
   - Area of a semicircle is \( \frac{1}{2} \pi r^2 \), where \( r = 1 \), so this gives us:
     \[
     \text{Area of semicircle} = \frac{1}{2} \pi \cdot 1^2 = \frac{\pi}{2}
     \]

2. **Triangle (from \( x = 1 \) to \( x = 3 \)):**
   - The triangle has a base of 2 (from \( x = 1 \) to \( x = 3 \)) and a height of 2.
   - Area of the triangle is:
     \[
     \text{Area of triangle} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 2 = 2
     \]

Thus, the total area from \( x = 0 \) to \( x = 3 \) is:
\[
\frac{\pi}{2} + 2
\]

Now, using the fundamental theorem, we have:
\[
g(3) = g(0) + \int_0^3 g'(x) \, dx = 5 + \left( \frac{\pi}{2} + 2 \right)
\]
\[
g(3) = 5 + \frac{\pi}{2} + 2 = 7 + \frac{\pi}{2}
\]

### Finding \( g(-2) \):
1. From \( x = 0 \) to \( x = -2 \), we again integrate the graph of \( g'(x) \).
2. This corresponds to two line segments, both below the x-axis.

#### Step-by-step:
1. **Line segment from \( x = 0 \) to \( x = -1 \):**
   - This is a straight line with a slope of -1, forming a triangle with base 1 and height 1 below the x-axis.
   - The area of the triangle is:
     \[
     \text{Area} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}
     \]
   - Since it's below the x-axis, the area is \( -\frac{1}{2} \).

2. **Line segment from \( x = -1 \) to \( x = -2 \):**
   - This is a triangle with a base of 1 and height of 1 below the x-axis.
   - The area is again \( \frac{1}{2} \), but negative:
     \[
     \text{Area} = -\frac{1}{2}
     \]

Thus, the total area from \( x = 0 \) to \( x = -2 \) is:
\[
- \frac{1}{2} - \frac{1}{2} = -1
\]

Now, using the fundamental theorem:
\[
g(-2) = g(0) + \int_0^{-2} g'(x) \, dx = 5 + (-1) = 4
\]

### Final answers:
- \( g(3) = 7 + \frac{\pi}{2} \)
- \( g(-2) = 4 \)

Would you like further details or have any other questions? Here are some related questions you might explore:

1. What if the semicircle had a different radius? How would that affect \( g(3) \)?
2. How do we apply the Fundamental Theorem of Calculus to other types of graphs?
3. Can we find the maximum or minimum values of \( g(x) \) using this graph?
4. What happens if we reverse the limits of integration?
5. How would the calculation change if \( g'(x) \) had discontinuities?

**Tip**: When integrating graphically, always pay attention to the areas above and below the x-axis to determine the net contribution.

The function g is defined and differentiable on the closed interval [-7, 5] and satisfies g(0) = 5. The graph of y = g'(x), the derivative of g, consists of a semicircle and three line segments. Find g(3) and g(-2).

Learn how to find g(3) and g(-2) by integrating the derivative g'(x) from a graph. This problem covers geometric areas under the curve and applies the Fundamental Theorem of Calculus, suitable for undergraduate calculus students.

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