## Task B-2.5.

Determine the minimum value of the function $f: \mathbb{R} \backslash\left\{-\frac{1}{2}\right\} \rightarrow \mathbb{R}, f(x)=\frac{12 x^{2}+8 x+4}{(2 x+1)^{2}}$. For which $x$ does the function $f$ achieve its minimum?
## First Solution.

Write the function $f$ in the form:

$$
\begin{aligned}
f(x) & =\frac{12 x^{2}+8 x+4}{(2 x+1)^{2}}=\frac{3(2 x+1)^{2}-4 x+1}{(2 x+1)^{2}}=3+\frac{-4 x+1}{(2 x+1)^{2}}=3+\frac{-2(2 x+1)+3}{(2 x+1)^{2}}= \\
& =3-\frac{2}{2 x+1}+\frac{3}{(2 x+1)^{2}}
\end{aligned}
$$

With the substitution $\frac{1}{2 x+1}=t$, we consider the function $g(t)=3 t^{2}-2 t+3$.

The minimum value of the function $g$ is $m=\frac{4 a c-b^{2}}{4 a}=\frac{36-4}{12}=\frac{32}{12}=\frac{8}{3}$, and it is achieved at $t=-\frac{b}{2 a}=\frac{1}{3}$.

The minimum value of the function $f$ is equal to the minimum value of the function $g$.

Thus, the minimum value of the function $f$ is $\frac{8}{3}$.

Since the function $g$ achieves its minimum at $t=\frac{1}{3}$, from $\frac{1}{2 x+1}=\frac{1}{3}$ we get that the function $f$ achieves its minimum at $x=1$.