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To solve this problem, let's go through it step-by-step.

### Given Function
The function given is:
\[
g(t) = -3t^2 + 5t - 4
\]

We need to:
1. **Find the intervals on which the function is increasing or decreasing.**
2. **Identify any local extreme values (maxima or minima).**

### Step 1: Find the Derivative of \( g(t) \)
To analyze the intervals of increase and decrease, we start by finding the first derivative, \( g'(t) \), because it helps identify critical points where the slope changes, which in turn tells us where the function is increasing or decreasing.

\[
g(t) = -3t^2 + 5t - 4
\]
Differentiating with respect to \( t \):
\[
g'(t) = -6t + 5
\]

### Step 2: Set the Derivative Equal to Zero
To find the critical points, set \( g'(t) = 0 \) and solve for \( t \):
\[
-6t + 5 = 0
\]
\[
6t = 5
\]
\[
t = \frac{5}{6}
\]

So, \( t = \frac{5}{6} \) is the critical point.

### Step 3: Determine the Sign of \( g'(t) \) Around the Critical Point
Now, we need to determine whether the function is increasing or decreasing on either side of \( t = \frac{5}{6} \).

1. **For \( t < \frac{5}{6} \):** Choose a test point, say \( t = 0 \).
   \[
   g'(0) = -6(0) + 5 = 5 > 0
   \]
   So, \( g'(t) > 0 \) when \( t < \frac{5}{6} \), indicating that \( g(t) \) is **increasing** on this interval.

2. **For \( t > \frac{5}{6} \):** Choose a test point, say \( t = 1 \).
   \[
   g'(1) = -6(1) + 5 = -1 < 0
   \]
   So, \( g'(t) < 0 \) when \( t > \frac{5}{6} \), indicating that \( g(t) \) is **decreasing** on this interval.

### Step 4: Conclusion for Part (a)
- **Increasing interval:** \( \left(-\infty, \frac{5}{6}\right) \)
- **Decreasing interval:** \( \left(\frac{5}{6}, \infty\right) \)

### Step 5: Identify the Local Extremum for Part (b)
Since the function changes from increasing to decreasing at \( t = \frac{5}{6} \), this point is a local maximum.

#### Find the value of \( g(t) \) at \( t = \frac{5}{6} \):
\[
g\left(\frac{5}{6}\right) = -3\left(\frac{5}{6}\right)^2 + 5\left(\frac{5}{6}\right) - 4
\]
Calculating each term:
1. \( \left(\frac{5}{6}\right)^2 = \frac{25}{36} \)
2. \( -3 \cdot \frac{25}{36} = -\frac{75}{36} = -\frac{25}{12} \)
3. \( 5 \cdot \frac{5}{6} = \frac{25}{6} \)
4. Convert \( \frac{25}{6} \) to \( \frac{50}{12} \) to combine terms.

Now:
\[
g\left(\frac{5}{6}\right) = -\frac{25}{12} + \frac{50}{12} - \frac{48}{12} = \frac{-23}{12}
\]

So, the local maximum value of \( g(t) \) is \( -\frac{23}{12} \), occurring at \( t = \frac{5}{6} \).

### Final Answers
- **(a) Intervals of Increase and Decrease:**
  - Increasing on \( \left(-\infty, \frac{5}{6}\right) \)
  - Decreasing on \( \left(\frac{5}{6}, \infty\right) \)

- **(b) Local Extreme Values:**
  - Local maximum of \( -\frac{23}{12} \) at \( t = \frac{5}{6} \).

Would you like further explanation on any part of this solution, or do you have additional questions?

### Related Questions
1. How can we determine concavity for \( g(t) \) and possible inflection points?
2. What are other ways to confirm if a critical point is a maximum or minimum?
3. How would we find the absolute maximum and minimum over a closed interval?
4. What is the significance of the second derivative in analyzing the function's behavior?
5. How can we apply these methods to functions with higher degrees?

### Tip
When analyzing functions, always start by finding the first derivative to identify critical points. Then use sign analysis or the second derivative test to classify these points accurately.

Find the open intervals on which the function is increasing and those on which it is decreasing. Identify the function's local extreme values, if any, saying where they occur. Given: g(t) = -3t^2 + 5t - 4

Learn how to determine the intervals of increase and decrease and find local extreme values for the function g(t) = -3t^2 + 5t - 4. This solution includes step-by-step differentiation, finding critical points, and using the first derivative test for maxima and minima. Suitable for high school calculus students.

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