The problem involves analyzing the function $g(x) = \int_0^x f(t) \, dt$, where $f(t)$ is the given graph. Let's approach each question step-by-step:

(a) Local Maxima and Minima of $g(x)$

To find the local maxima and minima of $g(x)$, we need to understand where the derivative $g'(x) = f(x)$ changes sign (i.e., where $f(x)$ crosses the $x$-axis or peaks).

1. Local Minima ( $x_{\text{min}}$ ):

   • Local minima of $g(x)$ occur where $f(x)$ changes from negative to positive (crosses the $x$-axis from below).
   • Observing the graph of $f(t)$:
     • The smaller $x$-value for a local minimum seems to be around $t = 10$.
     • The larger $x$-value for a local minimum is around $t = 30$.

   Thus, $x_{\text{min}} = 10 \quad \text{and} \quad x_{\text{min}} = 30$

2. Local Maxima ( $x_{\text{max}}$ ):

   • Local maxima of $g(x)$ occur where $f(x)$ changes from positive to negative.
   • Observing the graph of $f(t)$:
     • The smaller $x$-value for a local maximum appears around $t = 5$.
     • The larger $x$-value for a local maximum appears around $t = 35$.

   Thus, $x_{\text{max}} = 5 \quad \text{and} \quad x_{\text{max}} = 35$

(b) Absolute Maximum of $g(x)$

The absolute maximum of $g(x)$ occurs at the point where the cumulative area under $f(t)$ from $t = 0$ to $x$ reaches its highest positive value.

Looking at the graph of $f(t)$, the area under the curve seems to accumulate the most by around $t = 5$. Therefore: $x = 5$

(c) Concave Downward Intervals for $g(x)$

The concavity of $g(x)$ is determined by the sign of $f'(x)$. $g(x)$ is concave downward where $f(x)$ is decreasing.

From the graph, $f(t)$ is decreasing in the interval approximately from $t = 5$ to $t = 15$ and again from $t = 30$ to $t = 40$.

Thus, the interval where $g(x)$ is concave downward is: $(5, 15) \cup (30, 40)$

(d) Sketching the Graph of $g(x)$

For sketching $g(x)$, use the following observations:

• $g(x)$ starts at 0 when $x = 0$.
• $g(x)$ increases where $f(x)$ is positive and decreases where $f(x)$ is negative.
• Local maxima and minima of $g(x)$ correspond to where $f(x)$ crosses the $x$-axis or reaches peaks and troughs.
• Concave up and concave down behavior corresponds to intervals where $f(x)$ is increasing or decreasing, respectively.

Would you like more detailed steps on any part of this solution or assistance with the graph of $g(x)$?

Follow-Up Questions:

1. How can we determine concavity from the second derivative in general?
2. Why does the integral of a function $f(t)$ provide insight into the accumulation of areas under $f(t)$?
3. How does the Fundamental Theorem of Calculus relate the function $g(x)$ and its derivative?
4. Why does $g(x)$ reach a local maximum where $f(x)$ crosses from positive to negative?
5. How would we find the exact value of $g(x)$ at specific points if $f(t)$ were given as a formula?

Tip:

When analyzing the integral function $g(x) = \int_0^x f(t) \, dt$, keep in mind that $g(x)$ reflects the net area under $f(t)$ from 0 to $x$. Positive areas increase $g(x)$, and negative areas decrease it.