Let \(f(x) = \frac{4}{{x + 2}},{\text{ }}x \ne - 2{\text{ and }}g(x) = x - 1\).

If \(h = g \circ f\) , find

(a)     h(x) ;

(b)     \({h^{ - 1}}(x)\) , where \({h^{ - 1}}\) is the inverse of h.

(a)     \(h(x) = g\left( {\frac{4}{{x + 2}}} \right)\)     (M1)

\( = \frac{4}{{x + 2}} - 1\,\,\,\,\,\left( { = \frac{{2 - x}}{{2 + x}}} \right)\)     A1

(b)     METHOD 1

\(x = \frac{4}{{y + 2}} - 1\,\,\,\,\,\)(interchanging x and y)     M1

Attempting to solve for y     M1

\((y + 2)(x + 1) = 4\,\,\,\,\,\left( {y + 2 = \frac{4}{{x + 1}}} \right)\)     (A1)

\({h^{ - 1}}(x) = \frac{4}{{x + 1}} - 2\,\,\,\,\,(x \ne - 1)\)     A1     N1

METHOD 2

\(x = \frac{{2 - y}}{{2 + y}}\,\,\,\,\,\)(interchanging x and y)     M1

Attempting to solve for y     M1

\(xy + y = 2 - 2x\,\,\,\,\,\left( {y(x + 1) = 2(1 - x)} \right)\)     (A1)

\({h^{ - 1}}(x) = \frac{{2(1 - x)}}{{x + 1}}\,\,\,\,\,(x \ne - 1)\)     A1     N1

Note: In either METHOD 1 or METHOD 2 rearranging first and interchanging afterwards is equally acceptable.

[6 marks]

This question was generally well done, with very few candidates calculating \(f \circ g\) rather than \(g \circ f\).