Find the maximum value of $f(x)=\dfrac{1}{4 x^{2}+2 x+1}$.

To find the maximum value of $f(x) = \dfrac{1}{4x^{2} + 2x + 1}$, we need to find the minimum value of the denominator $g(x) = 4x^{2} + 2x + 1$.

The expression $g(x) = 4x^{2} + 2x + 1$ is a quadratic in the form $ax^{2} + bx + c$ where $a = 4$, $b = 2$, and $c = 1$. Since $a > 0$, the quadratic has a minimum value at $x = -\dfrac{b}{2a}$.

$x = -\dfrac{2}{2 \times 4} = -\dfrac{2}{8} = -\dfrac{1}{4}$.

Substituting $x = -\dfrac{1}{4}$ into $g(x)$ to find the minimum value:

$g\left(-\dfrac{1}{4}\right) = 4\left(-\dfrac{1}{4}\right)^{2} + 2\left(-\dfrac{1}{4}\right) + 1 = 4\left(\dfrac{1}{16}\right) - \dfrac{1}{2} + 1 = \dfrac{1}{4} - \dfrac{1}{2} + 1 = \dfrac{1 - 2 + 4}{4} = \dfrac{3}{4}$.

Since the minimum value of the denominator is $\dfrac{3}{4}$, the maximum value of $f(x)$ is the reciprocal of this minimum value:

$f_{max} = \dfrac{1}{3/4} = \dfrac{4}{3}$.

Answer: $\dfrac{4}{3}$