Answer: (i) \(gf(x)=\dfrac{8x-5}{12x-10}\). (ii) \(g^{-1}(x)=\dfrac{x+1}{3x-2}\). (iii) \(x=2\).
(i) Since \(f(x)=4x-3\),
\(gf(x)=g(f(x)).\)
Substitute \(4x-3\) into \(g\):
\(g(4x-3) = \frac{2(4x-3)+1}{3(4x-3)-1}.\)
Simplify:
\(gf(x) = \frac{8x-6+1}{12x-9-1} = \frac{8x-5}{12x-10}.\)
(ii) Let
\(y=\frac{2x+1}{3x-1}.\)
Rearrange to make \(x\) the subject:
\(y(3x-1)=2x+1.\)
So
\(3xy-y=2x+1.\)
Collect the terms containing \(x\):
\(3xy-2x=y+1.\)
Factorise:
\(x(3y-2)=y+1.\)
Therefore
\(x=\frac{y+1}{3y-2}.\)
Replacing \(y\) by \(x\),
\(g^{-1}(x)=\frac{x+1}{3x-2}.\)
(iii) First find \(fg(x)=f(g(x))\):
\(f(g(x)) = 4\left(\frac{2x+1}{3x-1}\right)-3.\)
Use a common denominator:
\(f(g(x)) = \frac{8x+4-3(3x-1)}{3x-1} = \frac{8x+4-9x+3}{3x-1} = \frac{7-x}{3x-1}.\)
Now solve \(fg(x)=x-1\):
\(\frac{7-x}{3x-1}=x-1.\)
Multiply by \(3x-1\):
\(7-x=(x-1)(3x-1).\)
Expand and rearrange:
\(7-x=3x^2-4x+1,\)
so
\(3x^2-3x-6=0.\)
Divide by \(3\):
\(x^2-x-2=0.\)
Factorise:
\((x+1)(x-2)=0.\)
Since the functions are defined for \(x\gt 1\), \(x=-1\) is not valid. Hence
\(\boxed{x=2}.\)
