Find the points of local maxima and minima of following functions
`(i) f(x) =(x-1)(x-2)^(2) " "(ii) f(x) =-(x-1)^(3) (x+1)^(2)`
`(iii) f(x) =xe^x`

Answer

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

1. **Differentiate the function** to find its first derivative.
2. **Set the first derivative equal to zero** to find critical points.
3. **Determine the nature of each critical point** (whether it is a maximum or minimum) using the second derivative test or the first derivative test.

Let's solve each function step by step.

### (i) \( f(x) = (x-1)(x-2)^2 \)

**Step 1: Differentiate the function.**

Using the product rule:
\[ f'(x) = (x-1) \cdot \frac{d}{dx}[(x-2)^2] + (x-2)^2 \cdot \frac{d}{dx}[(x-1)] \]
Calculating the derivatives:
\[ \frac{d}{dx}[(x-2)^2] = 2(x-2) \]
\[ \frac{d}{dx}[(x-1)] = 1 \]
Substituting back:
\[ f'(x) = (x-1)(2(x-2)) + (x-2)^2(1) \]
\[ = 2(x-1)(x-2) + (x-2)^2 \]
\[ = (x-2)(2(x-1) + (x-2)) \]
\[ = (x-2)(2x - 2 + x - 2) = (x-2)(3x - 4) \]

**Step 2: Set the first derivative equal to zero.**
\[ f'(x) = 0 \implies (x-2)(3x-4) = 0 \]
This gives us:
\[ x - 2 = 0 \implies x = 2 \]
\[ 3x - 4 = 0 \implies x = \frac{4}{3} \]

**Step 3: Determine the nature of critical points.**

To determine whether these points are maxima or minima, we can use the second derivative test or analyze the first derivative around these points.

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

**Step 1: Differentiate the function.**

Using the product rule:
\[ f'(x) = -[(x-1)^3 \cdot \frac{d}{dx}[(x+1)^2] + (x+1)^2 \cdot \frac{d}{dx}[(x-1)^3]] \]
Calculating the derivatives:
\[ \frac{d}{dx}[(x+1)^2] = 2(x+1) \]
\[ \frac{d}{dx}[(x-1)^3] = 3(x-1)^2 \]
Substituting back:
\[ f'(x) = -[(x-1)^3 \cdot 2(x+1) + (x+1)^2 \cdot 3(x-1)^2] \]
Factoring out common terms:
\[ = -[(x-1)^2(x+1)(2(x-1) + 3(x+1))] \]
\[ = -[(x-1)^2(x+1)(5x + 1)] \]

**Step 2: Set the first derivative equal to zero.**
\[ f'(x) = 0 \implies (x-1)^2(x+1)(5x+1) = 0 \]
This gives us:
\[ x - 1 = 0 \implies x = 1 \quad (\text{double root}) \]
\[ x + 1 = 0 \implies x = -1 \]
\[ 5x + 1 = 0 \implies x = -\frac{1}{5} \]

**Step 3: Determine the nature of critical points.**

Again, we can use the second derivative test or analyze the first derivative around these points.

### (iii) \( f(x) = xe^x \)

**Step 1: Differentiate the function.**

Using the product rule:
\[ f'(x) = e^x + xe^x = e^x(1 + x) \]

**Step 2: Set the first derivative equal to zero.**
\[ f'(x) = 0 \implies e^x(1 + x) = 0 \]
Since \( e^x \) is never zero, we have:
\[ 1 + x = 0 \implies x = -1 \]

**Step 3: Determine the nature of the critical point.**

We can use the second derivative test:
\[ f''(x) = e^x(1 + x) + e^x = e^x(2 + x) \]
Evaluating at \( x = -1 \):
\[ f''(-1) = e^{-1}(2 - 1) = \frac{1}{e} > 0 \]
Since \( f''(-1) > 0 \), it indicates that \( x = -1 \) is a local minimum.

### Summary of Critical Points
1. For \( f(x) = (x-1)(x-2)^2 \): Critical points at \( x = 2 \) and \( x = \frac{4}{3} \).
2. For \( f(x) = -(x-1)^3(x+1)^2 \): Critical points at \( x = 1 \), \( x = -1 \), and \( x = -\frac{1}{5} \).
3. For \( f(x) = xe^x \): Critical point at \( x = -1 \) (local minimum).