Find the intervals in which the functions :
(i) `f(x)=x^(3)+2x^(2)-1`
(ii) `30-24x+15x^(2)-2x^(3)` are strictly decreasing.

Answer

To find the intervals in which the given functions are strictly decreasing, we need to follow these steps:

### (i) For the function \( f(x) = x^3 + 2x^2 - 1 \):

1. **Find the derivative**:
\[ f'(x) = \frac{d}{dx}(x^3 + 2x^2 - 1) = 3x^2 + 4x \]

2. **Set the derivative less than zero**:
To find where the function is strictly decreasing, we need to solve:
\[ f'(x) < 0 \]
This means:
\[ 3x^2 + 4x < 0 \]

3. **Factor the derivative**:
\[ x(3x + 4) < 0 \]

4. **Find critical points**:
Set each factor to zero:
\[ x = 0 \quad \text{and} \quad 3x + 4 = 0 \implies x = -\frac{4}{3} \]

5. **Test intervals**:
The critical points divide the number line into intervals: \( (-\infty, -\frac{4}{3}), (-\frac{4}{3}, 0), (0, \infty) \).
- Choose test points from each interval:
- For \( x = -2 \): \( f'(-2) = -6 < 0 \) (decreasing)
- For \( x = -1 \): \( f'(-1) = 0 < 0 \) (decreasing)
- For \( x = 1 \): \( f'(1) = 7 > 0 \) (increasing)

6. **Conclusion**:
The function \( f(x) \) is strictly decreasing in the interval:
\[ \left(-\frac{4}{3}, 0\right) \]

### (ii) For the function \( g(x) = 30 - 24x + 15x^2 - 2x^3 \):

1. **Find the derivative**:
\[ g'(x) = \frac{d}{dx}(30 - 24x + 15x^2 - 2x^3) = -24 + 30x - 6x^2 \]

2. **Set the derivative less than zero**:
To find where the function is strictly decreasing, we need to solve:
\[ g'(x) < 0 \]
This means:
\[ -6x^2 + 30x - 24 < 0 \]

3. **Factor the derivative**:
Dividing by -6 (and reversing the inequality):
\[ x^2 - 5x + 4 > 0 \]
Factoring gives:
\[ (x - 4)(x - 1) > 0 \]

4. **Find critical points**:
Set each factor to zero:
\[ x = 1 \quad \text{and} \quad x = 4 \]

5. **Test intervals**:
The critical points divide the number line into intervals: \( (-\infty, 1), (1, 4), (4, \infty) \).
- Choose test points from each interval:
- For \( x = 0 \): \( g'(0) = -24 < 0 \) (decreasing)
- For \( x = 2 \): \( g'(2) = -6 < 0 \) (decreasing)
- For \( x = 5 \): \( g'(5) = 6 > 0 \) (increasing)

6. **Conclusion**:
The function \( g(x) \) is strictly decreasing in the intervals:
\[ (-\infty, 1) \cup (4, \infty) \]

### Final Answer:
- The function \( f(x) \) is strictly decreasing in \( \left(-\frac{4}{3}, 0\right) \).
- The function \( g(x) \) is strictly decreasing in \( (-\infty, 1) \cup (4, \infty) \).