If the function `f(x)=x^4+b x^2+8x+1` has a horizontal tangent and a point of inflection for the same value of `x` then the value of `b` is equal to `-1` (b) 1 (c) 6 (d) `-6`

Options

A. `-2`
B. `-6`
C. 6
D. 3

Explanation

To solve the problem, we need to find the value of \( b \) for the function \( f(x) = x^4 + bx^2 + 8x + 1 \) such that it has a horizontal tangent and a point of inflection at the same value of \( x \).

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

\[ f'(x) = \frac{d}{dx}(x^4 + bx^2 + 8x + 1) = 4x^3 + 2bx + 8 \]

### Step 2: Set the first derivative equal to zero for horizontal tangent
For a horizontal tangent, we set \( f'(x) = 0 \):

\[ 4x^3 + 2bx + 8 = 0 \tag{1} \]

### Step 3: Find the second derivative \( f''(x) \)
Next, we find the second derivative \( f''(x) \) to determine the point of inflection:

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

### Step 4: Set the second derivative equal to zero for point of inflection
For a point of inflection, we set \( f''(x) = 0 \):

\[ 12x^2 + 2b = 0 \tag{2} \]

### Step 5: Solve for \( b \) from equation (2)
From equation (2), we can express \( b \) in terms of \( x \):

\[ 2b = -12x^2 \implies b = -6x^2 \tag{3} \]

### Step 6: Substitute \( b \) from equation (3) into equation (1)
Now we substitute \( b = -6x^2 \) into equation (1):

\[ 4x^3 + 2(-6x^2)x + 8 = 0 \]
\[ 4x^3 - 12x^3 + 8 = 0 \]
\[ -8x^3 + 8 = 0 \]
\[ -8x^3 = -8 \implies x^3 = 1 \implies x = 1 \]

### Step 7: Substitute \( x = 1 \) back into equation (3) to find \( b \)
Now we substitute \( x = 1 \) back into equation (3) to find \( b \):

\[ b = -6(1^2) = -6 \]

### Conclusion
Thus, the value of \( b \) is:

\[ \boxed{-6} \]