<s>[INST] 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 
 [/INST] 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}
</s>