Two functions g and h are defined on the set…

Question 1 Report

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 \(x = -\frac{1}{2}\).

Answer

Answer Details

\(g:x\to x^{2}-2\) and \(h:x\to\dfrac{1}{x+2}\).

(a) Inverse of h. Let \(y=\dfrac{1}{x+2}\). Then

\[x+2=\frac{1}{y}\;\Rightarrow\;x=\frac{1}{y}-2\]

So, replacing \(y\) by \(x\),

\[h^{-1}(x)=\frac{1}{x}-2=\frac{1-2x}{x}\]

(b) \(g\circ h\) at \(x=-\dfrac12\). First evaluate \(h\):

\[h\left(-\tfrac12\right)=\frac{1}{-\tfrac12+2}=\frac{1}{\tfrac32}=\frac{2}{3}\]

Then apply \(g\):

\[g\left(\tfrac23\right)=\left(\tfrac23\right)^{2}-2=\frac{4}{9}-2=\frac{4-18}{9}=-\frac{14}{9}\]

Hence \((g\circ h)\left(-\tfrac12\right)=-\dfrac{14}{9}\).

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 inve...