Functions: VJC 2024 Promo Q11

(a)

11(b)$0\in R_f$ but $0\notin D_g$$R_f\nsubseteq D_g$, $gf$ does not exist.

11(c)

Every horizontal line $y=k$ cuts the graph of $y=g(x)$ at most once. Hence, $g$ is one-to-one. Inverse of $g$ exists.

$$y=-\ln(x-1)\qquad e^{-y}=x-1\qquad x=e^{-y}+1$$ $g^{-1}(x)=e^{-x}+1$

11(c)$R_{g^{-1}}=D_g=[2,\infty)$$R_{g^{-1}}$ be the new domain of $f$the graph in part (a), $R_{fg^{-1}}=\left(0,\dfrac{23}{3}\right]$

Alternatively

Sketch the graph $y=fg^{-1}(x)=\dfrac{5(e^{-x}+1)+13}{(e^{-x}+1)^2-1}$ for

$D_{g^{-1}}=R_g=(-\infty,0]$ and $R_{fg^{-1}}=\left(0,\dfrac{23}{3}\right]$