Functions

Question code: 1A68

Question 0204

Functions f{f} and g{g} are defined by
f:x↦16x+1for x∈R,x≠−16,g:x↦x2for x∈R,x>−8. \begin{align*} &f : x \mapsto \frac{1}{ 6 x + 1 } && \textrm{for } x \in \mathbb{R}, x \neq - \frac{1}{6}, \\ &g : x \mapsto x^2 && \textrm{for } x \in \mathbb{R}, x > -8. \end{align*}

(i)

Explain why one of the composite functions fg{fg} and gf{gf} does not exist. Give a definition, including the domain, of the composite that exists.
[3]

(ii)

Find f−1(x){f^{-1}(x)} and state the domain of f−1.{f^{-1}.}
[3]
Answer

(i)

gf{gf} does not exist because Rf=(−∞,0)∪(0,∞,)⊈[0,∞]=Dg.R_f = {\left( -\infty, 0 \right) \cup \left( 0, \infty, \right)} \not \subseteq {\left[ 0, \infty \right]} = D_g.
fg:x↦16x2+1,for x∈R,x>−8.\displaystyle fg : x \mapsto \frac{1}{ 6 x^2 + 1 }, \quad \allowbreak {\textrm{for } x \in \mathbb{R}, x > -8.}

(ii)

f−1(x)=161x−16.{\displaystyle f^{-1}(x) = \frac{1}{6} \frac{1}{ x } - \frac{1}{6}.}
Df−1=Rf=(−∞,0)∪(0,∞).{\displaystyle D_{ f^{-1} } = R_f = \left( -\infty, 0 \right) \cup \left( 0, \infty \right).}
Question code: 1A68