`f(x)=x^(4)-8x^(3)+22x^(2)-24x+20` has minimum value at x =

Options

A. 0
B. `-1`
C. 2
D. 1

Answer

To find the minimum value of the function \( f(x) = x^4 - 8x^3 + 22x^2 - 24x + 20 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^4 - 8x^3 + 22x^2 - 24x + 20) \]
Calculating the derivative term by term, we get:
\[ f'(x) = 4x^3 - 24x^2 + 44x - 24 \]

### Step 2: Set the first derivative to zero
To find the critical points, we set the first derivative equal to zero:

\[ 4x^3 - 24x^2 + 44x - 24 = 0 \]

### Step 3: Simplify the equation
We can divide the entire equation by 4 to simplify it:

\[ x^3 - 6x^2 + 11x - 6 = 0 \]

### Step 4: Factor the cubic equation
We will use the trial and error method to find the roots of the cubic polynomial. Testing \( x = 1 \):

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

Since \( x = 1 \) is a root, we can factor \( (x - 1) \) out of the cubic polynomial. We perform polynomial long division or synthetic division to divide \( x^3 - 6x^2 + 11x - 6 \) by \( (x - 1) \):

After division, we find:

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

### Step 5: Factor the quadratic equation
Now, we factor the quadratic \( x^2 - 5x + 6 \):

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

Thus, we have:

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

### Step 6: Find the critical points
Setting each factor to zero gives us the critical points:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 7: Determine the nature of the critical points
To determine whether these points are minima or maxima, we will find the second derivative:

\[ f''(x) = \frac{d}{dx}(4x^3 - 24x^2 + 44x - 24) = 12x^2 - 48x + 44 \]

Now we evaluate the second derivative at the critical points:

1. At \( x = 1 \):
\[ f''(1) = 12(1)^2 - 48(1) + 44 = 12 - 48 + 44 = 8 \quad (\text{positive, so minimum}) \]

2. At \( x = 2 \):
\[ f''(2) = 12(2)^2 - 48(2) + 44 = 48 - 96 + 44 = -4 \quad (\text{negative, so maximum}) \]

3. At \( x = 3 \):
\[ f''(3) = 12(3)^2 - 48(3) + 44 = 108 - 144 + 44 = 8 \quad (\text{positive, so minimum}) \]

### Conclusion
The function has minimum values at \( x = 1 \) and \( x = 3 \). Therefore, the minimum value occurs at:

\[ \text{Minimum value at } x = 1 \text{ and } x = 3 \]