Step 1

1) Given that,

\(f(f(x))=9x-8...\)(1)

Step 2

Since degree of given composition is 1 so degree of f is also 1.

Consider,

\(f(x)=ax+b\)

So

\((f \cdot f)(x)=f(f(x))\)

\(=f(ax+b)\)

\(=a(ax+b)+b (Since\ f(x)=ax+b)\)

\(\displaystyle={a}^{{{2}}}{x}+{a}{b}+{b}\)...(2)

Step 3

Using (1) and (2),

\(\displaystyle{9}{x}-{8}={a}^{{{2}}}{x}+{a}{b}+{b}\)

Compare the coefficient of x and constant term,

\(\displaystyle{a}^{{{2}}}={9}\Rightarrow{a}=\pm{3}\)

And

\(\displaystyle{a}{b}+{b}=-{8}\Rightarrow{b}=-{\frac{{{8}}}{{{a}+{1}}}}\)

When \(a=3, b=-2\)

When \(a=-3, b=4\)

So

When \(a=3\) and \(b=-2\ then\ f(x)=3x-2\)

When \(a=-3\) and \(b=4\ then\ f(x)=-3x+4\)

Step 4

Hence,

\(f(x)=3x-2\)

\(f(x)=-3x+4\)