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To determine the intervals on which the function \( f(x) = x^4 - 4x^3 + 7 \) is increasing or decreasing, and to find any local extreme values, we follow these steps:

### Step 1: Differentiate \( f(x) \) to find \( f'(x) \)
1. Given \( f(x) = x^4 - 4x^3 + 7 \).
2. Compute \( f'(x) \):
   \[
   f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(4x^3) + \frac{d}{dx}(7)
   \]
   \[
   f'(x) = 4x^3 - 12x^2
   \]

### Step 2: Find the critical points
1. Set \( f'(x) = 0 \):
   \[
   4x^3 - 12x^2 = 0
   \]
2. Factor out \( 4x^2 \):
   \[
   4x^2(x - 3) = 0
   \]
3. Solve for \( x \):
   \[
   x = 0 \quad \text{or} \quad x = 3
   \]
   So, the critical points are \( x = 0 \) and \( x = 3 \).

### Step 3: Determine the sign of \( f'(x) \) on intervals around the critical points
1. Consider the intervals \( (-\infty, 0) \), \( (0, 3) \), and \( (3, \infty) \).
2. Choose test points in each interval and substitute them into \( f'(x) = 4x^3 - 12x^2 \) to determine the sign:
   - For \( x = -1 \) (in \( (-\infty, 0) \)):
     \[
     f'(-1) = 4(-1)^3 - 12(-1)^2 = -4 - 12 = -16 \quad (\text{negative})
     \]
   - For \( x = 1 \) (in \( (0, 3) \)):
     \[
     f'(1) = 4(1)^3 - 12(1)^2 = 4 - 12 = -8 \quad (\text{negative})
     \]
   - For \( x = 4 \) (in \( (3, \infty) \)):
     \[
     f'(4) = 4(4)^3 - 12(4)^2 = 4 \cdot 64 - 12 \cdot 16 = 256 - 192 = 64 \quad (\text{positive})
     \]

### Step 4: Analyze intervals and identify local extrema
- On \( (-\infty, 0) \), \( f'(x) < 0 \), so \( f(x) \) is **decreasing**.
- On \( (0, 3) \), \( f'(x) < 0 \), so \( f(x) \) is **decreasing**.
- On \( (3, \infty) \), \( f'(x) > 0 \), so \( f(x) \) is **increasing**.

Since \( f(x) \) changes from decreasing to increasing at \( x = 3 \), there is a **local minimum at \( x = 3 \)**.

### Step 5: Find the local minimum value
Substitute \( x = 3 \) into \( f(x) \) to find the local minimum value:
\[
f(3) = (3)^4 - 4(3)^3 + 7 = 81 - 108 + 7 = -20
\]
Thus, the local minimum is at \( (3, -20) \).

### Summary
- **Increasing Interval**: \( (3, \infty) \)
- **Decreasing Intervals**: \( (-\infty, 0) \) and \( (0, 3) \)
- **Local Minimum**: \( (3, -20) \)

Would you like a more detailed explanation, or do you have any questions?

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### Related Questions:
1. How do you determine concavity and inflection points for \( f(x) \)?
2. What is the global minimum or maximum of \( f(x) = x^4 - 4x^3 + 7 \) on a closed interval?
3. How do we use the second derivative test to confirm local extrema?
4. Can we determine where \( f(x) \) is concave up or down?
5. What effect does changing the coefficient of \( x^3 \) have on the function’s extrema?

### Tip:
Always test points within each interval around critical points to determine whether the function is increasing or decreasing, as this helps to clearly establish the function's behavior on those intervals.

Find the open intervals of x-values on which f(x) = x^4 - 4x^3 + 7 is increasing and decreasing, as well as any local extreme values.

Explore how to determine the intervals on which the function f(x) = x^4 - 4x^3 + 7 is increasing or decreasing, including identifying local extrema. This detailed solution involves calculus concepts like derivatives and the first derivative test, suitable for high school students in Grades 11-12.

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