`f(x)=x^(4)-12x^(3)+52x^(2)-96x+48` has maximum value at x = …..

Options

A. `-4`
B. 2
C. 3
D. 4

Answer

To find the value of \( x \) at which the function \( f(x) = x^4 - 12x^3 + 52x^2 - 96x + 48 \) has a maximum value, we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
To find the critical points where the maximum or minimum values might occur, we first need to compute the first derivative of the function.

\[ f'(x) = \frac{d}{dx}(x^4 - 12x^3 + 52x^2 - 96x + 48) \]
Using the power rule, we differentiate each term:

\[ f'(x) = 4x^3 - 36x^2 + 104x - 96 \]

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

\[ 4x^3 - 36x^2 + 104x - 96 = 0 \]

### Step 3: Simplify the equation
To simplify, we can divide the entire equation by 4:

\[ x^3 - 9x^2 + 26x - 24 = 0 \]

### Step 4: Use the Rational Root Theorem or trial and error to find roots
We can try possible rational roots. Testing \( x = 2 \):

\[ 2^3 - 9(2^2) + 26(2) - 24 = 8 - 36 + 52 - 24 = 0 \]

Thus, \( x = 2 \) is a root.

### Step 5: Factor the polynomial
Now we can factor \( x - 2 \) out of \( x^3 - 9x^2 + 26x - 24 \) using synthetic division:

\[ \begin{array}{r|rrrr} 2 & 1 & -9 & 26 & -24 \\ & & 2 & -14 & 24 \\ \hline & 1 & -7 & 12 & 0 \\ \end{array} \]

This gives us:

\[ x^3 - 9x^2 + 26x - 24 = (x - 2)(x^2 - 7x + 12) \]

### Step 6: Factor the quadratic
Now we factor \( x^2 - 7x + 12 \):

\[ x^2 - 7x + 12 = (x - 3)(x - 4) \]

Thus, we have:

\[ f'(x) = (x - 2)(x - 3)(x - 4) = 0 \]

### Step 7: Find the critical points
The critical points are:

\[ x = 2, \quad x = 3, \quad x = 4 \]

### Step 8: Determine the nature of the critical points
To determine whether these points are maxima or minima, we will use the second derivative test.

### Step 9: Find the second derivative \( f''(x) \)
We differentiate \( f'(x) \):

\[ f''(x) = \frac{d}{dx}(4x^3 - 36x^2 + 104x - 96) = 12x^2 - 72x + 104 \]

### Step 10: Evaluate the second derivative at the critical points
1. **At \( x = 2 \)**:

\[ f''(2) = 12(2^2) - 72(2) + 104 = 48 - 144 + 104 = 8 \quad (\text{positive, minimum}) \]

2. **At \( x = 3 \)**:

\[ f''(3) = 12(3^2) - 72(3) + 104 = 108 - 216 + 104 = -4 \quad (\text{negative, maximum}) \]

3. **At \( x = 4 \)**:

\[ f''(4) = 12(4^2) - 72(4) + 104 = 192 - 288 + 104 = 8 \quad (\text{positive, minimum}) \]

### Conclusion
The function \( f(x) \) has a maximum value at \( x = 3 \).