Two functions g and h are defined on the set R of real numbers by (g : x to x^{2} – 2) and (h : x to frac{1}{x + 2}). Find :
(a) (h^{-1}), the inverse of h ;
(b) (g circ h), when (x = -frac{1}{2}).
Explanation
(a) (h(x) = frac{1}{x + 2}) ;
Let y = h(x).
(y = frac{1}{x + 2})
(frac{1}{y} = x + 2 implies x = frac{1}{y} – 2)
(therefore h^{-1} (x) = frac{1}{x} – 2)
(b) (g circ h(x) = g(h(x)))
(g(h(-frac{1}{2})) = g(frac{1}{-frac{1}{2} + 1}))
= (g(frac{1}{frac{3}{2}}))
= (g(frac{2}{3}))
= ((frac{2}{3})^{2} – 2)
= (frac{4}{9} – 2)
= (frac{-14}{9})