Two functions g and h are defined on the set R of real numbers by...

FURTHER MATHEMATICS
WAEC 2012

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) \(g \circ h(x) = g(h(x))\)

\(g(h(-\frac{1}{2})) = g(\frac{1}{-\frac{1}{2} + 1})\)

= \(g(\frac{1}{\frac{3}{2}})\)

= \(g(\frac{2}{3})\)

= \((\frac{2}{3})^{2} - 2\)

= \(\frac{4}{9} - 2\)

= \(\frac{-14}{9}\)



Post an Explanation Or Report an Error
If you see any wrong question or answer, please leave a comment below and we'll take a look. If you doubt why the selected answer is correct or need additional more details? Please drop a comment or Contact us directly. Your email address will not be published. Required fields are marked *
Add Math
Don't want to keep filling in name and email whenever you make a contribution? Register or login to make contributing easier.