Let's go through each part of the problem.

Problem Interpretation

You’re given a function $g(x) = \int_0^x f(t) \, dt$, where $f$ is represented by the provided graph. We need to evaluate $g(x)$ at specified values by approximating the area under $f$ from $t = 0$ to $t = x$. Since the graph of $f$ does not have an explicit formula, we'll approximate these areas by calculating the areas of geometric shapes (like triangles and rectangles) formed under the curve.

Step-by-Step Solution

(a) Evaluate $g(x)$ for $x = 0, 5, 10, 15, 20, 25,$ and $30$.

1. $g(0) = 0$
   Since $g(0)$ represents the integral from $0$ to $0$, it has no area under the curve, so $g(0) = 0$.

2. $g(5)$
   From $0$ to $5$, we have a triangle below the x-axis with a base of 5 and height of approximately -4.
   Area $= \frac{1}{2} \times 5 \times (-4) = -10$.
   Thus, $g(5) = -10$.

3. $g(10)$
   From $5$ to $10$, we have a triangle above the x-axis with a base of 5 and height of approximately 4.
   Area $= \frac{1}{2} \times 5 \times 4 = 10$.
   Adding this to $g(5)$, we get $g(10) = -10 + 10 = 0$.

4. $g(15)$
   From $10$ to $15$, we have a triangle below the x-axis with a base of 5 and height of -3.
   Area $= \frac{1}{2} \times 5 \times (-3) = -7.5$.
   Thus, $g(15) = 0 - 7.5 = -7.5$.

5. $g(20)$
   From $15$ to $20$, we have a rectangle above the x-axis with a base of 5 and height of 3.
   Area $= 5 \times 3 = 15$.
   Adding this to $g(15)$, we get $g(20) = -7.5 + 15 = 7.5$.

6. $g(25)$
   From $20$ to $25$, we have a triangle above the x-axis with a base of 5 and height of 5.
   Area $= \frac{1}{2} \times 5 \times 5 = 12.5$.
   Thus, $g(25) = 7.5 + 12.5 = 20$.

7. $g(30)$
   From $25$ to $30$, we have a triangle below the x-axis with a base of 5 and height of -5.
   Area $= \frac{1}{2} \times 5 \times (-5) = -12.5$.
   Thus, $g(30) = 20 - 12.5 = 7.5$.

(b) Estimate $g(35)$.

Using the midpoint method, we’ll approximate the area from $30$ to $35$ with the average height of $f$ in that interval, which appears to be around 2.
Approximate area from $30$ to $35 = 5 \times 2 = 10$.
Thus, $g(35) \approx 7.5 + 10 = 17.5$.

(c) Identify where $g$ has a maximum and minimum value.

Based on the calculations in part (a):

• Minimum value occurs at $x = 15$ with $g(15) = -7.5$.
• Maximum value occurs at $x = 25$ with $g(25) = 20$.

Final Answers

• (a)
  $g(0) = 0$
  $g(5) = -10$
  $g(10) = 0$
  $g(15) = -7.5$
  $g(20) = 7.5$
  $g(25) = 20$
  $g(30) = 7.5$

• (b)
  $g(35) \approx 17.5$

• (c)
  Minimum at $x = 15$ with $g(15) = -7.5$
  Maximum at $x = 25$ with $g(25) = 20$

Would you like more details on any of these steps?

Additional Related Questions

1. How would changing the heights of the peaks in $f(t)$ affect the values of $g(x)$?
2. What does the function $g(x)$ represent in terms of accumulated area?
3. How could we approximate $g(x)$ if $f(t)$ had irregular shapes instead of straight-line segments?
4. What happens to $g(x)$ if $f(t)$ were shifted vertically?
5. Can we determine the concavity of $g(x)$ from the graph of $f(x)$?

Tip

When working with integrals that represent areas, visualize each segment carefully and remember that areas below the x-axis are negative.