Question: > 
> Let $f(x)=a(x-2)(x-b)$, where $a,b\in R$ and $a\ne0$. Also, $f(f(x))=a^3\left(x^2-(2+b)x+2b-\frac2a\right)\left(x^2-(2+b)x+2b-\frac ba\right)$, $a\ne0$ has exactly one real zeroes $5$. Find the minima and maxima of $f(x)$.
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$f(f(x))$ is a quartic equation. So, it would have $4$ roots. Its real root should occur in pair because complex roots occur in pair. So, I failed to understand the meaning of 'exactly one real zero' in the question. Does that mean its real roots are equal and they are equal to $5$?

Since $f(x)$ has a zero at $2$, $f(f(2))=f(0)=2ab$. But even RHS of $f(f(x))$ is coming out to be $2ab$ at $x=2$.

Putting $x=5$ in $f(f(x))$ gives an equation in $a^2$ and $b^2$. Don't know what to do with it.

Also, $f(x)=a(x^2-(2+b)x+2b)$

$f'(x)=a(2x-2-b)=0\implies x=\frac{2+b}2$ is the minima or maxima.

How to proceed next? Do the critical points of $f(x)$ tell us anything about $f(f(x))$?

Answer: 
Let $q>3$. Then there exists a nonzero element $a\in \Bbb F\_q$ with $a^2-1\neq 0$. Let $A=\begin{pmatrix} 1 & -x \cr 0 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} a & 0 \cr 0 & a^{-1} \end{pmatrix}$ in $SL(2,q)$. Then
\begin{align\*}
[A,B] & = ABA^{-1}B^{-1} \\
 & =\begin{pmatrix} 1 & (a^2-1)x \cr 0 & 1 \end{pmatrix}
\end{align\*}
So every element of the corresponding transvection group
is a commutator. Since the transvections generate
$SL(2,q)$, this is enough to show that $SL(2,q)$ is perfect.
I might approach it via Bruhat decomposition. Let $B$ be the group of upper triangular matrices in $G$. It is the semidirect product of the group $T$ of diagonal matrices in $G$ and the group $N$ of upper triangular matrices with $1$s on the diagonal. First, show that every element of $B$ is a product of commutators in $G$.

Next, use the fact that $G$ is the disjoint union of the double cosets $BwB$, where $w$ ranges through a set of permutation matrices in $G$. This is called the *Bruhat decomposition*. For example, $G = \operatorname{SL}\_2$ is the disjoint union of $B = B1B$ and $BwB$, where

$$w = \begin{pmatrix} & 1 \\ -1 \end{pmatrix}.$$

Thus you just have to show that $B$ and every such $w$ is a product of commutators.