Find the points of local maxima or minima of following functions
(i) `f(x) = (x-1)^(3) (x+2)^(2)`
(ii) `f(x) =x^(3) +x^(2) +x+1.`

Answer

To find the points of local maxima or minima for the given functions, we will follow these steps:

### Part (i): \( f(x) = (x - 1)^3 (x + 2)^2 \)

1. **Differentiate the function**:
We will use the product rule for differentiation. Let \( u = (x - 1)^3 \) and \( v = (x + 2)^2 \).
\[ f'(x) = u'v + uv' \]
First, we find \( u' \) and \( v' \):
\[ u' = 3(x - 1)^2, \quad v' = 2(x + 2) \]
Now substituting back:
\[ f'(x) = 3(x - 1)^2 (x + 2)^2 + (x - 1)^3 \cdot 2(x + 2) \]

2. **Factor the derivative**:
We can factor out common terms:
\[ f'(x) = (x - 1)^2 (x + 2) [3(x + 2) + 2(x - 1)] \]
Simplifying the expression inside the brackets:
\[ 3(x + 2) + 2(x - 1) = 3x + 6 + 2x - 2 = 5x + 4 \]
Thus, we have:
\[ f'(x) = (x - 1)^2 (x + 2)(5x + 4) \]

3. **Set the derivative to zero**:
To find critical points, we set \( f'(x) = 0 \):
\[ (x - 1)^2 = 0 \quad \Rightarrow \quad x = 1 \]
\[ (x + 2) = 0 \quad \Rightarrow \quad x = -2 \]
\[ (5x + 4) = 0 \quad \Rightarrow \quad x = -\frac{4}{5} \]

4. **Determine the nature of critical points**:
We will use the first derivative test. We check the sign of \( f'(x) \) around the critical points:
- For \( x < -2 \): Choose \( x = -3 \) → \( f'(-3) > 0 \) (increasing)
- For \( -2 < x < -\frac{4}{5} \): Choose \( x = -1 \) → \( f'(-1) < 0 \) (decreasing)
- For \( -\frac{4}{5} < x < 1 \): Choose \( x = 0 \) → \( f'(0) < 0 \) (decreasing)
- For \( x > 1 \): Choose \( x = 2 \) → \( f'(2) > 0 \) (increasing)

From this, we conclude:
- Local maximum at \( x = -2 \)
- Local minimum at \( x = -\frac{4}{5} \)

### Part (ii): \( f(x) = x^3 + x^2 + x + 1 \)

1. **Differentiate the function**:
\[ f'(x) = 3x^2 + 2x + 1 \]

2. **Set the derivative to zero**:
We solve \( 3x^2 + 2x + 1 = 0 \) using the discriminant:
\[ D = b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot 1 = 4 - 12 = -8 \]
Since the discriminant is negative, there are no real roots.

3. **Conclusion**:
Since \( f'(x) \) does not equal zero for any real \( x \), the function has no local maxima or minima.

### Summary of Results:
- For \( f(x) = (x - 1)^3 (x + 2)^2 \):
- Local maximum at \( x = -2 \)
- Local minimum at \( x = -\frac{4}{5} \)

- For \( f(x) = x^3 + x^2 + x + 1 \):
- No local maxima or minima.