Find the intervals of monotonicity for the following fucntions.
`(i) (x^(4))/(4)+(x^(3))/(3)-3x^(2)+5 " "(ii) log_(3)^(2) x+ log_(3)x`

Answer

To find the intervals of monotonicity for the given functions, we will follow these steps:

### Part (i): \( f(x) = \frac{x^4}{4} + \frac{x^3}{3} - 3x^2 + 5 \)

1. **Differentiate the function**:
\[ f'(x) = \frac{d}{dx}\left(\frac{x^4}{4}\right) + \frac{d}{dx}\left(\frac{x^3}{3}\right) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(5) \]
\[ f'(x) = x^3 + x^2 - 6x \]

2. **Set the derivative equal to zero**:
\[ f'(x) = x^3 + x^2 - 6x = 0 \]
Factor out \( x \):
\[ x(x^2 + x - 6) = 0 \]

3. **Factor the quadratic**:
\[ x^2 + x - 6 = (x - 2)(x + 3) \]
Thus, we have:
\[ f'(x) = x(x - 2)(x + 3) = 0 \]

4. **Find the critical points**:
The critical points are:
\[ x = 0, \quad x = 2, \quad x = -3 \]

5. **Determine the intervals**:
The critical points divide the number line into intervals:
- \( (-\infty, -3) \)
- \( (-3, 0) \)
- \( (0, 2) \)
- \( (2, \infty) \)

6. **Test each interval**:
- For \( x < -3 \) (e.g., \( x = -4 \)):
\[ f'(-4) = (-4)((-4) - 2)((-4) + 3) = -4 \cdot -6 \cdot -1 = -24 \quad (\text{decreasing}) \]
- For \( -3 < x < 0 \) (e.g., \( x = -1 \)):
\[ f'(-1) = (-1)((-1) - 2)((-1) + 3) = -1 \cdot -3 \cdot 2 = 6 \quad (\text{increasing}) \]
- For \( 0 < x < 2 \) (e.g., \( x = 1 \)):
\[ f'(1) = (1)(1 - 2)(1 + 3) = 1 \cdot -1 \cdot 4 = -4 \quad (\text{decreasing}) \]
- For \( x > 2 \) (e.g., \( x = 3 \)):
\[ f'(3) = (3)(3 - 2)(3 + 3) = 3 \cdot 1 \cdot 6 = 18 \quad (\text{increasing}) \]

7. **Conclusion**:
- Increasing on \( (-3, 0) \) and \( (2, \infty) \)
- Decreasing on \( (-\infty, -3) \) and \( (0, 2) \)

### Part (ii): \( f(x) = \log_3(x^2) + \log_3(x) \)

1. **Rewrite the function**:
\[ f(x) = \log_3(x^2) + \log_3(x) = \log_3(x^2 \cdot x) = \log_3(x^3) \]

2. **Differentiate the function**:
Using the chain rule:
\[ f'(x) = \frac{3}{x \ln(3)} \]

3. **Set the derivative equal to zero**:
Since \( \frac{3}{x \ln(3)} \) is never zero, we need to find where it is undefined, which occurs at \( x = 0 \). However, \( x \) must be positive for the logarithm to be defined.

4. **Determine the intervals**:
The function is defined for \( x > 0 \).

5. **Test the interval**:
- For \( x > 0 \), \( f'(x) > 0 \) (since both \( 3 \) and \( \ln(3) \) are positive).
- Therefore, \( f(x) \) is increasing for \( x > 0 \).

### Final Results:
- For \( f(x) = \frac{x^4}{4} + \frac{x^3}{3} - 3x^2 + 5 \):
- Increasing on \( (-3, 0) \) and \( (2, \infty) \)
- Decreasing on \( (-\infty, -3) \) and \( (0, 2) \)

- For \( f(x) = \log_3(x^2) + \log_3(x) \):
- Increasing on \( (0, \infty) \)