Let’s tackle the problem step by step to ensure all parts are addressed clearly.

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(a) Using the Fundamental Theorem of Calculus to Graph $g'(x)$:

The first part of the Fundamental Theorem of Calculus states:

$g'(x) = f(x)$

This means the derivative of $g(x)$ is equal to $f(x)$. Hence, $g'(x)$ is just the graph of $f(x)$.

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Observations About the Graphs:

1. The magnitude of $g'(t)$ at the point $(t, g'(t))$ is the slope of $f$ at the point $(t, f(t))$: This is false because $g'(t) = f(t)$, not the slope of $f$.
2. $g'(x)$ is the same as $f(x)$: This is true.
3. The area under $g'(x)$ is greater than the area under $f(x)$: This is false since $g'(x) = f(x)$, and their areas must be equivalent.
4. $g'(x)$ is equal to zero where $f(x)$ has a maximum or minimum: This is true because at these points, $f(x) = 0$, which is the derivative $g'(x) = f(x)$.
5. $g'(x)$ is the inverse of $f(x)$: This is false as $g'(x) = f(x)$.

Correct observation: $g'(x) = f(x)$, and $g'(x) = 0$ where $f(x)$ has a maximum or minimum.

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(b) Finding $g(3)$, $g'(3)$, and $g''(3)$:

Given: $g(x) = \int_{0}^{x} f(t) \, dt$

1. $g(3)$: To compute $g(3)$, integrate $f(t)$ from 0 to 3:

   • From $t = 0$ to $t = 1$, $f(t)$ is a straight line increasing from 0 to 2.5 (triangle area):
     $\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 1 \cdot 2.5 = 1.25$
   • From $t = 1$ to $t = 2$, $f(t)$ is a straight line increasing from 2.5 to 5 (trapezoid area):
     $\text{Area} = \frac{1}{2} \cdot (\text{sum of parallel sides}) \cdot \text{height} = \frac{1}{2} \cdot (2.5 + 5) \cdot 1 = 3.75$
   • From $t = 2$ to $t = 3$, $f(t)$ is a straight line decreasing from 5 to 3 (trapezoid area):
     $\text{Area} = \frac{1}{2} \cdot (5 + 3) \cdot 1 = 4$

   Adding these areas: $g(3) = 1.25 + 3.75 + 4 = 9$

2. $g'(3)$: From the Fundamental Theorem of Calculus: $g'(3) = f(3)$ From the graph, $f(3) = 3$.

3. $g''(3)$: Since $g''(x) = f'(x)$:

   • At $x = 3$, the slope of $f(x)$ is calculated from the straight line from $x = 2$ to $x = 4$, where $f(x)$ decreases linearly from 5 to 1: $f'(3) = \frac{1 - 5}{4 - 2} = -2$

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Final Answers:

1. $g(3) = 9$
2. $g'(3) = 3$
3. $g''(3) = -2$

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Let me know if you want further clarification or a visual representation of the solution.

Questions for Further Exploration:

1. What would $g(x)$ look like for $x > 3$?
2. How does $g''(x)$ change as $x$ varies?
3. Could the integral $\int_{a}^{b} f(t) dt$ be used to find $g(b) - g(a)$?
4. Why is $g(x)$ increasing or decreasing depending on the sign of $f(x)$?
5. Can you generalize $g(x)$ for other functions $f(t)$?

Tip: Always relate the derivative $g'(x)$ and the second derivative $g''(x)$ back to the slopes and curvature of the original function $f(x)$.