Example 3 Find the range of the following functions:
(1) $y=x+\sqrt{1-2 x}$;
(2) $y=\frac{10^{x}+10^{-x}}{10^{x}-10^{-x}}$;
(3) $y=\log _{0.5}\left(2 x-x^{2}+3\right)$;
(4) $y=\sqrt{x^{2}+2 x+2}+\sqrt{x^{2}-2 x+2}$;
(5) $y=|x+1|+|x-1|+|x+2|$;
(6) $y=\frac{x^{2}+4 x+3}{x^{2}+x-6}$;
(7) $y=x^{2}+\frac{1}{x} \quad\left(x \in \mathrm{R}^{+}\right)$.
Solution: (1) Let $\sqrt{1-2 x}=t \quad(t \geqslant 0)$, then
$$
x=\frac{1-t^{2}}{2}, y=1-\frac{1}{2}(t-1)^{2} \text {. }
$$

Since $t \geqslant 0$, we have $y=1-\frac{1}{2}(t-1)^{2} \leqslant 1$, which means the range of the function is $(-\infty, 1]$.
Explanation: By variable substitution, the problem is transformed into finding the range of a quadratic function. This can be done by analyzing the graph of the quadratic function or by using reverse reasoning to find the range. It is important to pay attention to the range of the variable after substitution.
(2) Since $y=\frac{10^{x}+10^{-x}}{10^{x}-10^{-x}}=\frac{10^{2 x}+1}{10^{2 x}-1}$, we have $10^{2 x}=\frac{y+1}{y-1}(y \neq 1)$, and the inverse function is $y=\frac{1}{2} \log \frac{x+1}{x-1}$.

From $\frac{x+1}{x-1}>0$, we know the domain of the inverse function is $(-\infty,-1) \cup(1,+\infty)$, so the range of the original function is $(-\infty,-1) \cup(1,+\infty)$.

Explanation: $1^{\circ}$. For functions with inverse functions, the range can be found using the domain of the inverse function. For functions without inverse functions, this idea can also be applied. For example, the range of $y=\frac{\sin x-1}{\sin x+2}$ can be found as follows: from $y=\frac{\sin x-1}{\sin x+2}$, we get $\sin x=\frac{1+2 y}{1-y}(y \neq 1)$. Since $|\sin x| \leqslant 1$, we have $\left|\frac{1+2 y}{1-y}\right| \leqslant 1$, solving which gives $-2 \leqslant y \leqslant 0$. This also shows that for Example 2, it is not necessary to find the inverse function; the range can be found by using the range of $10^{2 x}$, i.e., from $10^{2 x}=\frac{y+1}{y-1}>0$.
$2^{\circ}$. This example can also be simplified by separating the function expression, $y=\frac{10^{2 x}+1}{10^{2 x}-1}=1+\frac{2}{10^{2 x}+1}$. Let $10^{2 x}=t,(t>0)$, then $y=1+\frac{2}{t+1}$. The range can be easily found using the graph of the function $y=1+\frac{2}{x+1}$.
(3) Since $2 x-x^{2}+3=-(x-1)^{2}+4$, we have $0<2 x-x^{2}+3 \leqslant 4$. Since $0.5<1$, by the monotonicity of $y=\log _{0.1} x$, the range is $[-2,+\infty)$.

Explanation: Using the monotonicity of functions is also a common method for finding the range.
(4) Solution 1: $y=\sqrt{x^{2}+2 x+2}+\sqrt{x^{2}-2 x+2}$
$\geqslant 2 \sqrt[4]{\left(x^{2}+2\right)^{2}-4 x^{2}}$
$=2 \sqrt[4]{x^{4}+4} \geqslant 2 \sqrt[4]{4}=2 \sqrt{2}$ (equality holds when $x=0$), so the range is $[2 \sqrt{2},+\infty)$.

Solution 2: $y=\sqrt{(x+1)^{2}+1^{2}}+\sqrt{(x-1)^{2}+1^{2}}$ represents the sum of the distances from the moving point $P(x, 1)$ to the fixed points $A(1,0), B(-1,0)$. Hence, $y \geqslant 2 \sqrt{2}$, so the range is $[2 \sqrt{2},+\infty)$.

Explanation: $1^{\circ}$. When using the basic inequality to find the range of a function, it is important to check if the formula conditions are met and if the equality can be achieved. For example, $y=\sin x+\frac{2}{\sin x} \geqslant 2 \sqrt{2}$ is clearly incorrect.
$2^{\circ}$. Using geometric intuition is also a method for finding the range of a function.
$3^{\circ}$. This example can also be solved using complex numbers.
(5) $y=|x+1|+|x-1|+|x+2|$ represents the sum of the distances from the point $P$ with coordinate $x$ on the number line to the points $A(-1), B(1), C(-2)$. By observing the position of the moving point $P$, it is clear that the minimum value 3 is achieved when $P$ is at point $A(-1)$, so the range is $[3,+\infty)$.

Explanation: Generally, if there are points $A_{1}$,
Figure 2-1 $A_{2}, \cdots, A_{n}$, and $x_{A_{1}}<x_{A_{2}}<\cdots<x_{n}$, then $f(x)=\left|x-x_{A_{1}}\right|+\left|x-x_{A_{2}}\right|+$ $\cdots+\left|x-x_{A_{n}}\right|$.

When $n$ is odd, $x$ is at the middle point, and $f(x)$ reaches its minimum value; when $n$ is even, $x$ is between the two middle points, and $f(x)$ reaches its minimum value.
(6) Since $y=\frac{(x+3)(x+1)}{(x+3)(x-2)}=1+\frac{3}{x-2} \quad(x \neq-3)$, we have $y \neq 1$, and $y \neq \frac{2}{5}$, so the range is $\left(-\infty, \frac{2}{5}\right) \cup\left(\frac{2}{5}, 1\right) \cup(1,+\infty)$.

Explanation: If the discriminant method is used, it is not easy to notice that $y \neq \frac{2}{5}$, which would expand the range.
(7) Since $y=x^{2}+\frac{1}{x}=x^{2}+\frac{1}{2 x}+\frac{1}{2 x}$, and $3 \sqrt[3]{x^{2} \cdot \frac{1}{2 x} \cdot \frac{1}{2 x}}=$ $=\frac{3 \sqrt[3]{2}}{2}$, equality holds when $x^{2}=\frac{1}{2 x}$, i.e., $x=\frac{\sqrt[3]{4}}{2}$, so the range is $\left[\frac{3 \sqrt[3]{2}}{2},+\infty\right)$.