Let's write down the inverse of the function $f:(-\infty,-2) \rightarrow \mathbf{R}, x \mapsto 2 x^{2}+8 x+7$.
I. Solution. The quadratic function $\tilde{f}: \mathbf{R} \rightarrow \mathbf{R}, \tilde{f}(x)=2 x^{2}+8 x+7$ extended to real numbers is strictly decreasing on the interval $(-\infty,-2)$. Indeed, for example, after completing the square, it is evident that $\tilde{f}$ attains its minimum at $x_{0}=-2$, so for every $x$ in the domain of $f$, there is exactly one $f(x)$, and for every $f(x)$ function value, there is exactly one $x$ value: the function $f$ is bijective, hence the inverse $f^{-1}$ exists (Figure 1).

To determine the inverse relationship, we need to express the value of $x$ in terms of $y$ from the equation

$$
y=2 x^{2}+8 x+7
$$

for $x$ values less than -2. It is convenient to use the quadratic formula, so after rearranging to zero,

$$
x_{1,2}=-2 \pm \sqrt{\frac{1+y}{2}}
$$

From this, it is clear that only for $y \geq-1$ does there exist an $x$ such that $\tilde{f}(x)=y$, the range of $f$ is the interval $(-1,+\infty)$, which is the domain of $f^{-1}$. However, for such $y$ values, there are two $x$ values for which $\tilde{f}(x)=y$, we need to decide which of these is in the range of $f^{-1}$, i.e., in the domain of $f$, the interval $(-\infty,-2)$.

$x_{1}$ and $x_{2}$ are symmetrically placed around -2 (Figure 2). Of the two numbers, the smaller one is $x_{1}=-2-\sqrt{\frac{1+y}{2}}$.

$$
f_{1}(x)=x+2 ; \quad f_{2}(x)=-x ; \quad f_{3}(x)=x^{2} ; \quad f_{4}(x)=2 x-1
$$

Since the complete square form of the function $f$ is $f(x)=2[-(x+2)]^{2}-1$, then

$$
f: x \stackrel{f_{7}}{\mapsto} x+2 \stackrel{f_{2}}{\mapsto}-(x+2) \stackrel{f_{3}}{\mapsto}[-(x+2)]^{2} \stackrel{f_{4}}{\mapsto} 2[-(x+2)]^{2}-1,
$$

i.e., $f=f_{4} \circ f_{3} \circ f_{2} \circ f_{1}$.

From the components, it can be seen that they are bijective, and their inverses are respectively

$$
f_{1}^{-1}(x)=-2+x ; \quad f_{2}^{-1}(x)=-x ; \quad f_{3}^{-1}(x)=\sqrt{x}, \quad x>0 ; \quad f_{4}^{-1}=\frac{x+1}{2}
$$

Thus, the inverse of $f$ is $f^{-1}=f_{1}^{-1} \circ f_{2}^{-1} \circ f_{3}^{-1} \circ f_{4}^{-1}$, i.e.,

$$
f^{-1}: x \stackrel{f_{4}^{-1}}{\mapsto} \frac{x+1}{2} \stackrel{f_{3}^{-1}}{\mapsto} \sqrt{\frac{x+1}{2}} \stackrel{f_{2}^{-1}}{\mapsto}-\sqrt{\frac{x+1}{2}} \stackrel{f_{\mapsto}^{-1}}{\mapsto}-2+\left(-\sqrt{\frac{x+1}{2}}\right)
$$

so $f^{-1}(x)=-2-\sqrt{\frac{x+1}{2}}$. The domain of the function is the range of $f$, the interval $(-1 ;+\infty)$.

Remark. Many incorrect and incomplete solutions were received for the problem. Ignoring the superficiality of the terminology - several reflected the function, not the graph - the most glaring error is the formal, incorrect result of the quadratic equation in the first solution, $-2 \pm \sqrt{\frac{1+x}{2}}$ as the inverse function. Reviewing the solutions to the problem is a good opportunity for all our readers to reflect on concepts related to functions: domain, range, composite function, equation, graph, inverse, etc.