Find the intevals in which the following functionis decreasing. `f(x) =x^(4)-8x^(3)+22x^(2)-24x+21`

Answer

To find the intervals in which the function \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 21 \) is decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
We start by finding the first derivative \( f'(x) \) of the function.

\[ f'(x) = \frac{d}{dx}(x^4 - 8x^3 + 22x^2 - 24x + 21) \]
Using the power rule, we differentiate each term:

\[ f'(x) = 4x^3 - 24x^2 + 44x - 24 \]

### Step 2: Factor the derivative
Next, we will factor \( f'(x) \). We can factor out a common factor of 4:

\[ f'(x) = 4(x^3 - 6x^2 + 11x - 6) \]

Now, we will find the roots of the cubic polynomial \( x^3 - 6x^2 + 11x - 6 \) using trial and error or synthetic division. Testing \( x = 1 \):

\[ 1^3 - 6(1^2) + 11(1) - 6 = 1 - 6 + 11 - 6 = 0 \]

So, \( x = 1 \) is a root. We can factor \( x - 1 \) out of the cubic polynomial:

Using synthetic division, we divide \( x^3 - 6x^2 + 11x - 6 \) by \( x - 1 \):

\[ \begin{array}{r|rrrr} 1 & 1 & -6 & 11 & -6 \\ & & 1 & -5 & 6 \\ \hline & 1 & -5 & 6 & 0 \\ \end{array} \]

This gives us:

\[ x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6) \]

Now we can factor \( x^2 - 5x + 6 \):

\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

Thus, we have:

\[ f'(x) = 4(x - 1)(x - 2)(x - 3) \]

### Step 3: Determine where the derivative is less than or equal to zero
To find where the function is decreasing, we need to solve:

\[ f'(x) < 0 \]

This means we need to analyze the sign of \( 4(x - 1)(x - 2)(x - 3) \).

### Step 4: Identify critical points and test intervals
The critical points are \( x = 1, 2, 3 \). We will test the intervals created by these points:

1. \( (-\infty, 1) \)
2. \( (1, 2) \)
3. \( (2, 3) \)
4. \( (3, \infty) \)

We will pick test points from each interval:

- For \( x < 1 \) (e.g., \( x = 0 \)):
\[ f'(0) = 4(0 - 1)(0 - 2)(0 - 3) = 4(-1)(-2)(-3) = -24 < 0 \]

- For \( 1 < x < 2 \) (e.g., \( x = 1.5 \)):
\[ f'(1.5) = 4(1.5 - 1)(1.5 - 2)(1.5 - 3) = 4(0.5)(-0.5)(-1.5) = 4 \cdot 0.5 \cdot 0.5 \cdot 1.5 = 1.5 > 0 \]

- For \( 2 < x < 3 \) (e.g., \( x = 2.5 \)):
\[ f'(2.5) = 4(2.5 - 1)(2.5 - 2)(2.5 - 3) = 4(1.5)(0.5)(-0.5) = -7.5 < 0 \]

- For \( x > 3 \) (e.g., \( x = 4 \)):
\[ f'(4) = 4(4 - 1)(4 - 2)(4 - 3) = 4(3)(2)(1) = 24 > 0 \]

### Step 5: Conclusion
From our tests, we find that \( f'(x) < 0 \) on the intervals:

- \( (-\infty, 1) \)
- \( (2, 3) \)

Thus, the function \( f(x) \) is decreasing in the intervals:

\[ (-\infty, 1) \cup (2, 3) \]