Let's determine the minimum and maximum value of the expression $\sqrt{x-2}+\sqrt{3-x}$.
Solution. The number under the square root cannot be negative, so $2 \leq x \leq 3$. The value of the square root is also non-negative, so it is sufficient to determine the smallest and largest values of $y^{2}=(\sqrt{x-2}+\sqrt{3-x})^{2}$.

Let $\sqrt{x-2}=u$ and $\sqrt{3-x}=v$, then according to the identity $(u+v)^{2}+(u-v)^{2}=2\left(u^{2}+v^{2}\right)$,

$$
(\sqrt{x-2}+\sqrt{3-x})^{2}+(\sqrt{x-2}-\sqrt{3-x})^{2}=2[(x-2)+(3-x)]=2
$$

After rearrangement, introducing the notation $\sqrt{x-2}-\sqrt{3-x}=z$:

$$
y^{2}=2-(\sqrt{x-2}-\sqrt{3-x})^{2}=2-z^{2}
$$

$y^{2}$ will be the largest when $z^{2}$ is the smallest, i.e., 0. This is satisfied when $x-2=3-x$, and from this $x=2.5$. At this point, $z=0, y^{2}=2$, so the largest value of $y$ is $\sqrt{2}$.

To determine the smallest value, consider the expression for $y^{2}$:

$$
y^{2}=(x-2)+(3-x)+2 \sqrt{(x-2)(3-x)}=1+2 \sqrt{(x-2)(3-x)}
$$

By the definition of the square root, the second term is non-negative, so $y^{2} \geq 1$. This is possible if the second term is 0, i.e., $x=2$ or $x=3$. At these points, the expression takes its smallest value, which is $\sqrt{1}=1$.

Remark. If we introduce the variable $t=2.5-x$, then $|t| \leq \frac{1}{2}$ and the expression is a positive even function of the new variable:

$$
\sqrt{x-2}+\sqrt{3-x}=\sqrt{\frac{1}{2}+t}+\sqrt{\frac{1}{2}-t}=f(t)
$$

$f$ and $f^{2}$ are simultaneously minimal and maximal, $f^{2}(t)=1+2 \sqrt{\frac{1}{4}-t^{2}}$, and this is a strictly decreasing function of $t$ in the interval $\left[0 ; \frac{1}{2}\right]$: $f^{2}(0)=2 \geq f^{2}(t) \geq f^{2}\left(\frac{1}{2}\right)=1$, so the smallest value of the expression is 1 (if $|t|=\frac{1}{2}$, i.e., $x=2$ or $x=3$), and the largest value is $\sqrt{2}$ (if $t=0$, i.e., $x=\frac{5}{2}$).