2007 H2 Mathematics Paper 1 Question 2

Functions

Answers

(i)

fg{fg} does not exist since Rg⊈Df{R_g \not \subseteq D_f}
gf:x↦1(x−3)2for x∈R,x≠3{gf: x \mapsto \frac{1}{(x-3)^2} \quad} \allowbreak {\textrm{for } x \in \mathbb{R}, x \neq 3}

(ii)

f−1(x)=3+1x{f^{-1}(x) = 3 + \frac{1}{x}}
Df−1=(−∞,0)∪(0,∞){D_{f^{-1}} = (-\infty, 0) \cup (0, \infty) }

Full solutions

(i)

Rg=[0,∞)Df=(−∞,3)∪(3,∞)\begin{align*} R_g &= [0, \infty) \\ D_f &= (-\infty, 3) \cup (3, \infty) \end{align*}
Hence the composite function fg{fg} does not exists since Rg⊈Df  ■{R_g \not \subseteq D_f \; \blacksquare}
gf(x)=g(1x−3)=(1x−3)2=1(x−3)2\begin{align*} gf(x) &= g\left( \frac{1}{x-3} \right) \\ &= \left( \frac{1}{x-3} \right)^2 \\ &= \frac{1}{(x-3)^2} \end{align*}
gf:x↦1(x−3)2for x∈R,x≠3  ■gf: x \mapsto \frac{1}{(x-3)^2} \quad \textrm{for } x \in \mathbb{R}, x \neq 3 \; \blacksquare

(ii)

y=1x−3x−3=1yx=3+1y\begin{gather*} y = \frac{1}{x-3} \\ x - 3 = \frac{1}{y} \\ x = 3 + \frac{1}{y} \end{gather*}
f−1(x)=3+1x  ■f^{-1}(x) = 3 + \frac{1}{x} \; \blacksquare
Df−1=Rf=(−∞,0)∪(0,∞)  ■\begin{align*} D_{f^{-1}} &= R_f \\ &= (-\infty, 0) \cup (0, \infty) \; \blacksquare \end{align*}