An equation of tangent line at an inflection point of `f(x) = x^(4) - 6x^(3) + 12x^(2) - 8x + 3` is

Options

A. `y= 3x +4`
B. `y=4`
C. `y = 3x +2`
D. none of these

Answer

To find the equation of the tangent line at an inflection point of the function \( f(x) = x^4 - 6x^3 + 12x^2 - 8x + 3 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
The first derivative of the function gives us the slope of the tangent line.

\[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(6x^3) + \frac{d}{dx}(12x^2) - \frac{d}{dx}(8x) + \frac{d}{dx}(3) \]

Calculating each term:

- \( \frac{d}{dx}(x^4) = 4x^3 \)
- \( \frac{d}{dx}(6x^3) = 18x^2 \)
- \( \frac{d}{dx}(12x^2) = 24x \)
- \( \frac{d}{dx}(8x) = 8 \)
- \( \frac{d}{dx}(3) = 0 \)

Putting it all together:

\[ f'(x) = 4x^3 - 18x^2 + 24x - 8 \]

### Step 2: Find the second derivative \( f''(x) \)
The second derivative helps us determine the inflection points.

\[ f''(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(18x^2) + \frac{d}{dx}(24x) - \frac{d}{dx}(8) \]

Calculating each term:

- \( \frac{d}{dx}(4x^3) = 12x^2 \)
- \( \frac{d}{dx}(18x^2) = 36x \)
- \( \frac{d}{dx}(24x) = 24 \)
- \( \frac{d}{dx}(8) = 0 \)

Putting it all together:

\[ f''(x) = 12x^2 - 36x + 24 \]

### Step 3: Set the second derivative to zero to find inflection points
To find the inflection points, we set \( f''(x) = 0 \):

\[ 12x^2 - 36x + 24 = 0 \]

Dividing the entire equation by 12:

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

Factoring the quadratic:

\[ (x - 1)(x - 2) = 0 \]

Thus, the inflection points are at \( x = 1 \) and \( x = 2 \).

### Step 4: Find the y-values at the inflection points
We will calculate the y-values of the function at these points.

1. For \( x = 1 \):

\[ f(1) = 1^4 - 6(1^3) + 12(1^2) - 8(1) + 3 = 1 - 6 + 12 - 8 + 3 = 2 \]

2. For \( x = 2 \):

\[ f(2) = 2^4 - 6(2^3) + 12(2^2) - 8(2) + 3 = 16 - 48 + 48 - 16 + 3 = 3 \]

Thus, the inflection points are \( (1, 2) \) and \( (2, 3) \).

### Step 5: Find the slope of the tangent line at the inflection points
Since we are looking for the tangent line at an inflection point, we can use either point. We will use \( x = 2 \) because it provides a clearer slope calculation.

Calculating \( f'(2) \):

\[ f'(2) = 4(2^3) - 18(2^2) + 24(2) - 8 \]

Calculating each term:

- \( 4(8) = 32 \)
- \( -18(4) = -72 \)
- \( 24(2) = 48 \)
- \( -8 = -8 \)

Putting it all together:

\[ f'(2) = 32 - 72 + 48 - 8 = 0 \]

### Step 6: Write the equation of the tangent line
Since the slope at \( x = 2 \) is \( 0 \), the tangent line is horizontal. The equation of a horizontal line is given by \( y = k \), where \( k \) is the y-value at the point of tangency.

Thus, the equation of the tangent line at the inflection point \( (2, 3) \) is:

\[ y = 3 \]

### Final Answer
The equation of the tangent line at an inflection point is:

\[ y = 3 \]