Originally Posted by

**acc100jt** If $\displaystyle h(x)=\frac{9-5x}{1-x}, x\neq1$

The question requires the function $\displaystyle h^{2}(x)$

I've worked out $\displaystyle h^{2}(x)$, which is given by

$\displaystyle h^{2}(x)=\frac{4x-9}{x-2}$

Obviously $\displaystyle h^{2}$ is undefined when $\displaystyle x=2$.

My question is whether $\displaystyle h^{2}$ is defined when $\displaystyle x=1$

Even though the expression $\displaystyle \frac{4x-9}{x-2}$ is defined when $\displaystyle x=1$, $\displaystyle h^{2}$ is not defined at $\displaystyle x=1$ since $\displaystyle h$ is not defined at $\displaystyle x=1$, am I correct?