17. $f(x)$ is a function defined on $\mathbf{R}$ that is odd, and satisfies the following conditions: (1) For any $x, y \in \mathbf{R}$, $f(x+y)=f(x)+f(y)$; (2) When $x>0$, $f(x)<0$, and $f(1)=-2$. Find the maximum and minimum values of $f(x)$ on the interval $[-2,3]$.
17. Since $f(x)$ is an odd function on $\mathbf{R}$, we first consider the case where $x \geqslant 0$. Let $x_{1}>x_{2} \geqslant 0$, then $f\left(x_{1}\right)=$ $f\left(x_{1}-x_{2}+x_{2}\right)=f\left(x_{1}-x_{2}\right)+f\left(x_{2}\right)$. Since $x_{1}-x_{2}>0$, we have $f\left(x_{1}-x_{2}\right)<0$, thus $f\left(x_{1}\right)<$ $f\left(x_{2}\right)$, so $f(x)$ is decreasing on $[0,+\infty)$. By the property of odd functions, $f(x)$ is also decreasing on $(-\infty, 0]$, $f_{\max }=$ $f(-3)=-f(3)=-f(2+1)=-f(2)-f(1)=-3 f(1)=6, f_{\min }=f(3)=-6$