Functions 1.1 (Domain)

We note that $f(2)$ is undefined because $\displaystyle \lim_{x \to 2}\frac {x^3+2x-4}{x-2} = \infty$ . For $x \ne 2$ , we can write

$f(x) = \frac 1{\sqrt{1-\frac {x^3+2x-4}{x-2}}} = \sqrt{\frac {x-2}{2-x-x^3}} = \sqrt{\frac {x-2}{(1-x)(x^2+x+2)}}$

For $f(x)$ to be real, $\dfrac {x-2}{(1-x)(x^2+x+2)} \ge 0$ . Since $x^2 + x + 2 = \left(x+\dfrac 12\right)^2 + \dfrac 74 > 0$ , $f(x)$ is real when $\dfrac {x-2}{1-x} \ge 0$ or $x \in (1, 2]$ . But $f(2)$ is undefined. Therefore the domain of $f(x)$ is $x \in \boxed{(1,2)}$ .