1. ## Composite functions

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

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

I've worked out $h^{2}(x)$, which is given by
$h^{2}(x)=\frac{4x-9}{x-2}$

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

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

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

2. Originally Posted by acc100jt
If $h(x)=\frac{9-5x}{1-x}, x\neq1$

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

I've worked out $h^{2}(x)$, which is given by
$h^{2}(x)=\frac{4x-9}{x-2}$

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

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

Even though the expression $\frac{4x-9}{x-2}$ is defined when $x=1$, $h^{2}$ is not defined at $x=1$ since $h$ is not defined at $x=1$, am I correct?
I do not understand the notation that you have employed. What do you mean by $h^2(x)$ ?

3. compose h with itself, i.e.
$h^{2}(x)=h(h(x))$

4. Originally Posted by acc100jt
compose h with itself, i.e.
$h^{2}(x)=h(h(x))$
Oh. Well, then no, a restriction at x=1 must be made because of the way we have defined $h^2(x)$.

We can't get passed $h(x)$ through x=1. This means we'd never get to $h^2(x)$. This is subtle, but it is the case when dealing with functions.

5. Originally Posted by VonNemo19
Oh. Well, then no, a restriction at x=1 must be made because of the way we have defined $h^2(x)$.
You mean, "Well, then yes" since he was asking if it was correct that the function was NOT defined at x= 1!

We can't get passed $h(x)$ through x=1. This means we'd never get to $h^2(x)$. This is subtle, but it is the case when dealing with functions.