454. Given the function (rational fractional) $f(x)=\frac{2 x-3}{3 x^{2}-1}$. Find $f(-2) ; f(0) ; f(1)$.
Solution. $f(-2)=\frac{2(-2)-3}{3(-2)^{2}-1}=-\frac{7}{11} ; f(0)=3 ; f(1)=-\frac{1}{2}$. The function $f(x)$ is undefined at points $x_{1}=\frac{1}{\sqrt{3}}$ and $x_{2}=-\frac{1}{\sqrt{3}}$, as the denominator becomes zero at these points.