11.3. Find the domain and range of the function

$$
y=\sqrt{x^{2}+3 x+2}+\frac{1}{\sqrt{2 x^{2}+6 x+4}}
$$
Answer. Domain of definition $x-1$; range of values $y \geq \sqrt[4]{8}$.

Solution. The domain of definition is given by the inequality $x^{2}+3 x+2>0 \Leftrightarrow(x+2)(x+1)>0 \Leftrightarrow$ $x-1$. To find the range of values, let $t=\sqrt{x^{2}+3 x+2}$. Then $t>0$ and $y=t+\frac{1}{\sqrt{2}}$. We can investigate the function $y(t)$ using the derivative and verify that it decreases on the interval $\left(0 ; \frac{1}{\sqrt[4]{2}}\right)$ and increases on the interval $\left(\frac{1}{\sqrt[4]{2}} ;+\infty\right) ;$ the minimum value of $y$ is $y\left(\frac{1}{\sqrt[4]{2}}\right)=\frac{1}{\sqrt[4]{2}}+\frac{\sqrt[4]{2}}{\sqrt{2}}=\sqrt[4]{8}$, and there is no maximum value, since $y(t)>t$ and, therefore, $y(t)$ can take arbitrarily large positive values. Due to the continuity of the function $y(t)$, we obtain the range of values $(\sqrt[4]{8} ;+\infty)$.

We can also obtain the result without using the derivative by applying the inequality between the arithmetic mean and the geometric mean:
$y=t+\frac{1}{\sqrt{2} t} \geq 2 \sqrt{t \cdot \frac{1}{\sqrt{2}} t}=\sqrt[4]{8}$ (where equality is achieved when $t=\frac{1}{\sqrt{2} t}$, i.e., $t=\frac{1}{\sqrt[4]{2}}$ ).