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.