Recover the unknown function by undoing a logarithm · source reference

Printed question

Functions $f:x\mapsto1+\frac2{x-1}$, $x\in\mathbb R$, $x<1$; $g:x\mapsto\ln x$, $x\in\mathbb R$, $0<x<1$.

(i) Explain why composite $gf$ does not exist. [1]

(ii) Find an expression for $fg(x)$. Hence or otherwise find $(fg)^{-1}(0)$. [4]

(iii) Find $h(x)$ for each case: (a) $gh(x)=x$, [1] (b) $hg(x)=x^2+1$. [2]

[2018/DHS/Prelim/I/5]

Printed working

(i) $R_f=(-\infty, 1)\nsubseteq D_g=(0, 1)$. Thus $gf$ does not exist.

(ii) $fg(x)=f(\ln x)=1+2/(\ln x-1)=(\ln x+1)/(\ln x-1)$, $0<x<1$.

Method1: let $x=(fg)^{-1}(0)$. Then $fg(x)=0$ ⇒ $(\ln x+1)/(\ln x-1)=0$ ⇒ $\ln x+1=0$ ⇒ $\ln x=-1$ ⇒ $x=e^{-1}$. Thus $(fg)^{-1}(0)=e^{-1}$.

Method2 (find inverse directly): $y=(\ln x+1)/(\ln x-1)$ ⇒ $y\ln x-y=\ln x+1$ ⇒ $\ln x(y-1)=y+1$ ⇒ $\ln x=(y+1)/(y-1)$ ⇒ $x=e^{(y+1)/(y-1)}$.

$(fg)^{-1}(x)=e^{(x+1)/(x-1)}$ for $-1<x<1$, so $(fg)^{-1}(0)=e^{(0+1)/(0-1)}=e^{-1}$.

(iii)(a) $gh(x)=x\Rightarrow h(x)=g^{-1}(x)=e^x$, $x<0$.

(iii)(b) Method1: $hg(x)=x^2+1$ ⇒ $hg(g^{-1}(x))=(g^{-1}(x))^2+1$ ⇒ $h(x)=(e^x)^2+1=e^{2x}+1$.

Method2: $h(g(x))=x^2+1$ ⇒ $h(\ln x)=(e^{\ln x})^2+1$ ⇒ $h(x)=(e^x)^2+1=e^{2x}+1$, on $x<0$.

Teacher notes

The source graph shows $y=f(x)$, cuts $(-1, 0)$ and $(0, -1)$, with asymptotes $y=1$, $x=1$. No separate sidebar text is printed for this group; both inverse and missing-function methods are retained above.

Answer key

(i) $R_f\nsubseteq D_g$, does not exist. (ii) $e^{-1}$. (iii)(a) $e^x$. (iii)(b) $e^{2x}+1$.

Native diagrams