Find a polynomial f(x) of degree 5 which increases in the interval `(-infty, 2]` and `[6, infty)` and decreases in the interval [2,6]. Given that `f(0) = 3` and `f'(4)-0`.

Answer

To find a polynomial \( f(x) \) of degree 5 that meets the specified conditions, we will follow these steps:

### Step 1: Determine the behavior of \( f'(x) \)

Given that \( f(x) \) is increasing in the intervals \( (-\infty, 2] \) and \( [6, \infty) \), and decreasing in the interval \( [2, 6] \), we can infer the following about the first derivative \( f'(x) \):
- \( f'(x) > 0 \) for \( x \in (-\infty, 2) \) and \( x \in (6, \infty) \)
- \( f'(x) < 0 \) for \( x \in (2, 6) \)

This means that \( f'(x) \) must have roots at \( x = 2 \) and \( x = 6 \). Since it is decreasing between these points, we also need a critical point at \( x = 4 \) where \( f'(4) = 0 \).

### Step 2: Formulate \( f'(x) \)

Given the roots and the behavior of \( f'(x) \), we can write:
\[ f'(x) = k(x - 2)(x - 6)(x - 4)^2 \]
where \( k \) is a constant that we will determine later.

### Step 3: Expand \( f'(x) \)

Let's expand \( f'(x) \):
\[ f'(x) = k(x - 2)(x - 6)(x - 4)^2 \]
First, expand \( (x - 4)^2 \):
\[ (x - 4)^2 = x^2 - 8x + 16 \]
Now, expand \( (x - 2)(x - 6) \):
\[ (x - 2)(x - 6) = x^2 - 8x + 12 \]
Now, we can combine these:
\[ f'(x) = k(x^2 - 8x + 12)(x^2 - 8x + 16) \]
Next, we will expand this product.

### Step 4: Combine and simplify

Let’s multiply:
\[ f'(x) = k[(x^2 - 8x + 12)(x^2 - 8x + 16)] \]
Using the distributive property:
\[ = k[x^4 - 8x^3 + 16x^2 - 8x^3 + 64x^2 - 96x + 12x^2 - 96x + 192] \]
Combine like terms:
\[ = k[x^4 - 16x^3 + (16 + 64 + 12)x^2 - (96 + 96)x + 192] \]
\[ = k[x^4 - 16x^3 + 92x^2 - 192x + 192] \]

### Step 5: Integrate to find \( f(x) \)

Now we integrate \( f'(x) \):
\[ f(x) = \int f'(x) \, dx = k\left(\frac{x^5}{5} - 4x^4 + \frac{92}{3}x^3 - 96x^2 + 192x\right) + C \]

### Step 6: Use given conditions to find \( k \) and \( C \)

We know \( f(0) = 3 \):
\[ f(0) = k\left(0 - 0 + 0 - 0 + 0\right) + C = 3 \implies C = 3 \]
Now, we also know \( f'(4) = 0 \):
\[ f'(4) = k[(4 - 2)(4 - 6)(4 - 4)^2] = k[2 \cdot (-2) \cdot 0] = 0 \]
This condition is satisfied for any \( k \), so we can choose \( k = 1 \) for simplicity.

### Final Result

Thus, the polynomial is:
\[ f(x) = \frac{1}{5}x^5 - 4x^4 + \frac{92}{3}x^3 - 96x^2 + 192x + 3 \]