603. Find the greatest and least values of the function $f(x)=\sqrt{100-x^{2}}$ on the interval $[-6,8]$.
Solution. $1^{0}$. Find the critical points on the interval $[-6,8]$:

$$
f^{\prime}(x)=\frac{-2 x}{2 \sqrt{100-x^{2}}}=-\frac{x}{\sqrt{100-x^{2}}}
$$

On the considered interval, we have only one critical point $x=0$; and $f(0)=10$.

$2^{j}$. Calculate the values of the function at the endpoints of the interval:

$$
f(-6)=\sqrt{100-36}=8, f(8)=\sqrt{100-64}=6
$$

$3^{0}$. Denoting the maximum value by $M$ and the minimum value by $m$ on the interval, we get $M=f(0)=10, m=f(8)=6$; here the minimum value is achieved at the endpoint of the interval.

Remark. Finding the maximum and minimum values of functions can be simplified by using the following properties of continuous functions:

1) if the function $y=f(x)$ is continuous and increasing on the interval $[a, b]$, then $m=f(a)$ and $M=f(b)$;
2) if the function $y=f(x)$ is continuous and decreasing on the interval $[a, b]$, then $m=f(b)$ and $M=f(a)$;
3) if the function $y=f(x)$, continuous on the interval $[a, b]$, has only one maximum point $x_{0}$ (and no minimum points) on this interval, then the maximum value on this interval is $M=f\left(x_{0}\right)$;
4) if the function $y=f(x)$, continuous on the interval $[a, b]$, has only one minimum point $x_{0}$ (and no maximum points) on this interval, then the minimum value on this interval is $m=f\left(x_{0}\right)$.

604-609. Find the maximum value $M$ and the minimum value $m$ of the following functions on the specified intervals: