Inverse function. A function $h$ is defined by $h:x\rightarrow 2-\frac{a}{x}$, where $x\neq 0$ and $a$ is a constant. Given $\frac{1}{2}h^2(2)+h^{-1}(-1)=-1$, find the possible values of $a$. 
Can someone give me some hints? Thanks 
 A: Thank you for this problem.  I was not familiar with the notation $h^2(x)=h(h(x))$, but this is clearly what must be meant in this problem since interpreting $h^2(x)=(h(x))^2$, yields no real solutions.  
If $h(x)=2-\frac{a}{x}$, then we can solve for $h^{-1}(x)$ in the following manner.
$x=2-\frac{a}{h^{-1}(x)} \Rightarrow x-2=-\frac{a}{h^{-1}(x)} \Rightarrow \frac{1}{x-2}=-\frac{h^{-1}(x)}{a} \Rightarrow -\frac{a}{x-2}=h^{-1}(x)=\frac{a}{2-x}$
Therefore $h^{-1}(-1)=\frac{a}{3}$
Now if we interpret $h^2(x)$ to mean $h(h(x))$, then $$h^2(x)=h(2-\frac{a}{x})=2-\frac{a}{2-\frac{a}{x}}=2-\frac{a}{\frac{2x-a}{x}}=2-\frac{ax}{2x-a}=\frac{4x-2a-ax}{2x-a}$$
So that $\frac{1}{2}h^2(2)=\frac{1}{2}\frac{8-2a-2a}{4-a}=\frac{1}{2}\frac{8-4a}{4-a}=\frac{4-2a}{4-a}$
Plugging these values into your equation yields
$$\frac{4-2a}{4-a}+\frac{a}{3}=-1 \Rightarrow \frac{12-6a+4a-a^2}{12-3a}=-1 \Rightarrow 12-2a-a^2=3a-12$$
Bringing everything to one side gives us the following quadratic equation
$$a^2+5a-24=0$$
This equation is factorable and gives us $(a+8)(a-3)=0 \Rightarrow a=3$ and $a=-8$ are solutions.
A: Firstly, find the inverse function $h^{-1}$. I.e. let $y=2-a/x$ and hence $h^{-1}$ is defined as
\begin{align}x=2-\frac{a}{y} \ \ & \Longrightarrow \ \ \frac{a}{y}=2-x\\
&\Longrightarrow y=\frac{a}{2-x}.
\end{align}
Now, $h(2)=2-a/2$ and $h^{-1}(-1)=a/3$. You should be able to continue from here and get a quadratic equation in $a$.
Edit. Here is some extra help. Your equation can be evaluated with the information above:
\begin{align}
\frac{1}{2}h^2(2)+h^{-1}(-1)&=\frac{1}{2}(h(2))^2+h^{-1}(-1)\\
&=\frac{1}{2}\left(2-\frac{a}{2}\right)^2+\frac{a}{3}\\
\end{align}
