The maximum value of the function ` f(x) = (1)/( 4x^(2) + 2 x + 1)` is

Options

A. `(3)/(4)`
B. `-(3)/(4)`
C. `(4)/(3)`
D. `-(4)/(3)`

Answer

To find the maximum value of the function \( f(x) = \frac{1}{4x^2 + 2x + 1} \), we can follow these steps:

### Step 1: Identify the Denominator
The function is given as:
\[ f(x) = \frac{1}{4x^2 + 2x + 1} \]
To maximize \( f(x) \), we need to minimize the denominator \( g(x) = 4x^2 + 2x + 1 \).

### Step 2: Analyze the Quadratic Function
The function \( g(x) = 4x^2 + 2x + 1 \) is a quadratic function. Since the coefficient of \( x^2 \) (which is 4) is positive, the parabola opens upwards, indicating that it has a minimum point.

### Step 3: Find the Vertex of the Quadratic
The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula:
\[ x = -\frac{b}{2a} \]
Here, \( a = 4 \) and \( b = 2 \). Thus,
\[ x = -\frac{2}{2 \cdot 4} = -\frac{2}{8} = -\frac{1}{4} \]

### Step 4: Calculate the Minimum Value of \( g(x) \)
Now, substitute \( x = -\frac{1}{4} \) back into \( g(x) \) to find the minimum value:
\[ g\left(-\frac{1}{4}\right) = 4\left(-\frac{1}{4}\right)^2 + 2\left(-\frac{1}{4}\right) + 1 \]
Calculating each term:
\[ = 4 \cdot \frac{1}{16} - \frac{1}{2} + 1 \]
\[ = \frac{1}{4} - \frac{1}{2} + 1 \]
\[ = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{1 - 2 + 4}{4} = \frac{3}{4} \]

### Step 5: Find the Maximum Value of \( f(x) \)
Since the minimum value of \( g(x) \) is \( \frac{3}{4} \), the maximum value of \( f(x) \) is:
\[ f(x)_{\text{max}} = \frac{1}{g(x)_{\text{min}}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

### Conclusion
Thus, the maximum value of the function \( f(x) \) is:
\[ \boxed{\frac{4}{3}} \]