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 \(x = -\frac{1}{2}\).
(a) \(h^{-1}\), the inverse of h ;
(b) \(g \circ h\), when \(x = -\frac{1}{2}\).
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Correct Answer: Option n
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) \(g \circ h(x) = g(h(x))\)
\(g(h(-\frac{1}{2})) = g(\frac{1}{-\frac{1}{2} + 1})\)
= \(g(\frac{1}{\frac{3}{2}})\)
= \(g(\frac{2}{3})\)
= \((\frac{2}{3})^{2} - 2\)
= \(\frac{4}{9} - 2\)
= \(\frac{-14}{9}\)
(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) \(g \circ h(x) = g(h(x))\)
\(g(h(-\frac{1}{2})) = g(\frac{1}{-\frac{1}{2} + 1})\)
= \(g(\frac{1}{\frac{3}{2}})\)
= \(g(\frac{2}{3})\)
= \((\frac{2}{3})^{2} - 2\)
= \(\frac{4}{9} - 2\)
= \(\frac{-14}{9}\)