To find the maximum value of \( f(x) = \frac{1}{4x^2 + 2x + 1} \), we first differentiate \( f(x) \).
1. The function is \( f(x) = \frac{1}{4x^2 + 2x + 1} \).
2. Differentiate \( f(x) \) with respect to \( x \) using the chain rule:
\[
f'(x) = \frac{- (8x + 2)}{(4x^2 + 2x + 1)^2}
\]
3. Set \( f'(x) = 0 \) to find critical points. This occurs when the numerator is zero, i.e., \( 8x + 2 = 0 \), solving for \( x \), we get \( x = -\frac{1}{4} \).
4. To confirm whether this is a maximum, check the second derivative or use the nature of the function. We find that \( f(x) \) attains its maximum at \( x = -\frac{1}{4} \), and the maximum value is \( \frac{4}{3} \).