Find the values of x for which the following functions are maximum or minimum:
(i) ` x^(3)- 3x^(2) - 9x `
(ii) ` 4x^(3)-15x^(2)+12x+1`

Answer

To find the values of \( x \) for which the given functions are maximum or minimum, we will follow these steps:

### Part (i): Function \( f(x) = x^3 - 3x^2 - 9x \)

1. **Differentiate the function**:
\[ f'(x) = \frac{d}{dx}(x^3 - 3x^2 - 9x) = 3x^2 - 6x - 9 \]

2. **Set the derivative equal to zero**:
\[ 3x^2 - 6x - 9 = 0 \]

3. **Simplify the equation**:
Divide the entire equation by 3:
\[ x^2 - 2x - 3 = 0 \]

4. **Factor the quadratic**:
\[ (x - 3)(x + 1) = 0 \]

5. **Find the critical points**:
Setting each factor to zero gives:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]
\[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \]

### Part (ii): Function \( g(x) = 4x^3 - 15x^2 + 12x + 1 \)

1. **Differentiate the function**:
\[ g'(x) = \frac{d}{dx}(4x^3 - 15x^2 + 12x + 1) = 12x^2 - 30x + 12 \]

2. **Set the derivative equal to zero**:
\[ 12x^2 - 30x + 12 = 0 \]

3. **Simplify the equation**:
Divide the entire equation by 6:
\[ 2x^2 - 5x + 2 = 0 \]

4. **Factor the quadratic**:
\[ (2x - 1)(x - 2) = 0 \]

5. **Find the critical points**:
Setting each factor to zero gives:
\[ 2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2} \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Summary of Critical Points

- For \( f(x) = x^3 - 3x^2 - 9x \): The critical points are \( x = 3 \) and \( x = -1 \).
- For \( g(x) = 4x^3 - 15x^2 + 12x + 1 \): The critical points are \( x = \frac{1}{2} \) and \( x = 2 \).