Let `f(x)=(x-2)^2x^n, n in N ` Then f(x) has a minimum at

Options

A. `x=2 " for all " n in N `
B. x=2 ifn is odd
C. x=0 ifn is even
D. x=0 is if n is odd

Explanation

To find the minimum of the function \( f(x) = (x - 2)^2 x^n \), where \( n \in \mathbb{N} \), we will follow these steps:

### Step 1: Find the derivative \( f'(x) \)

We will use the product rule to differentiate \( f(x) \). The function can be expressed as:
\[ f(x) = (x - 2)^2 \cdot x^n \]

Using the product rule:
\[ f'(x) = u'v + uv' \]
where \( u = (x - 2)^2 \) and \( v = x^n \).

Calculating \( u' \) and \( v' \):
- \( u' = 2(x - 2) \)
- \( v' = nx^{n-1} \)

Thus, we have:
\[ f'(x) = 2(x - 2) \cdot x^n + (x - 2)^2 \cdot nx^{n-1} \]

### Step 2: Factor out common terms

Next, we can factor out the common terms from \( f'(x) \):
\[ f'(x) = (x - 2) \left[ 2x^n + (x - 2) \cdot nx^{n-1} \right] \]

### Step 3: Set the derivative equal to zero

To find the critical points, we set \( f'(x) = 0 \):
\[ (x - 2) \left[ 2x^n + (x - 2) \cdot nx^{n-1} \right] = 0 \]

This gives us two cases:
1. \( x - 2 = 0 \) which implies \( x = 2 \)
2. \( 2x^n + (x - 2) \cdot nx^{n-1} = 0 \)

### Step 4: Solve the second equation

For the second equation:
\[ 2x^n + (x - 2) \cdot nx^{n-1} = 0 \]

Rearranging gives:
\[ 2x^n = - (x - 2) \cdot nx^{n-1} \]
Dividing both sides by \( x^{n-1} \) (assuming \( x \neq 0 \)):
\[ 2x = -n(x - 2) \]
Expanding and rearranging:
\[ 2x = -nx + 2n \]
\[ (2 + n)x = 2n \]
Thus:
\[ x = \frac{2n}{2 + n} \]

### Step 5: Identify the critical points

The critical points are:
1. \( x = 2 \)
2. \( x = \frac{2n}{2 + n} \)

### Step 6: Determine the nature of the critical points

To determine whether these points are minima or maxima, we can evaluate the second derivative \( f''(x) \) or use the first derivative test. However, since we are asked for the minimum, we can analyze the behavior of \( f'(x) \) around these points.

### Conclusion

The function \( f(x) \) has a minimum at:
- \( x = 2 \)
- \( x = \frac{2n}{2 + n} \)