Compose through a restricted range, then use symmetry for area · source reference

Printed question

(a) Functions $f:x\mapsto(x-2)(x-4)$ for $x\in\mathbb R$, $x<4$; $g:x\mapsto x/(x+3)$ for $x\in\mathbb R$, $x\ne-3$.

Determine if $f$ is one-one, justifying your answer. [1]

Only one of $fg$ and $gf$ exists. Give a definition (including domain) of the composite that exists, and explain why the other does not exist. Find the range of the composite that exists. [6]

(b) $h:x\mapsto a(1+e^{-x})$, $x\in\mathbb R_0^+$, where $a$ is positive. Find $h^{-1}(x)$ and its domain. [3]

Given that $h$ and $h^{-1}$ intersect where $x=b$, and $\int_0^b h(x)\, dx=I$, find the area bounded by $h, h^{-1}$ and the axes in terms of $I, b$. [2]

[2010/TJC/Prelim/II/3]

Printed working

(a) Horizontal line $y=-0.5$ cuts the graph of $f$ twice ⇒ $f$ is not one-one.

$R_g=(-\infty, 1)\cup(1, \infty)$, $D_f=(-\infty, 4)$. Since $R_g\nsubseteq D_f$, $fg$ does not exist.

$R_f=[-1, \infty)$, $D_g=(-\infty, -3)\cup(-3, \infty)$. Since $R_f\subseteq D_g$, $gf$ exists.

$$gf(x)=\frac{(x-2)(x-4)}{(x-2)(x-4)+3}=\frac{x^2-6x+8}{x^2-6x+11}.$$

Definition $gf:x\mapsto(x^2-6x+8)/(x^2-6x+11)$ for $x\in\mathbb R$, $x<4$.

The source flow diagram prints $D_f=(-\infty, 4)\xrightarrow f R_f=[-1, \infty)\xrightarrow g R_{gf}=[-1, 1)$, then immediately concludes $R_{gf}=[-1/2, 1)$. The intermediate $-1$ is a source typo: $g(-1)=-1/2$, and correct range is $[-1/2, 1)$.

(b) The solution uses $x>0$: let $y=a(1+e^{-x})$ ⇒ $y/a-1=e^{-x}$ ⇒ $x=-\ln(y/a-1)$. Thus $h^{-1}(x)=-\ln(x/a-1)$, $D_{h^{-1}}=R_h=(a, 2a)$ under that interpretation.

Endpoint qualification

The question prints $\mathbb R_0^+$, normally including0, while solution explicitly uses $x>0$ and open endpoint $(a, 2a)$. If0 is included, inverse domain is $(a, 2a]$ and $h^{-1}(2a)=0$. Both interpretations retain the same area formula.

Area bounded by $h, h^{-1}$ and axes

$$\begin{aligned}&=2\times(\text{area bounded by }h\text{ and }y=x)\quad\text{by symmetry about }y=x\\ &=2\times(\text{shaded area }I-\text{blue right-angled triangle area})\\ &=2(I-\tfrac12b^2)=2I-b^2.\end{aligned}$$

Teacher notes

The horizontal-line test must include the whole printed domain $x<4$.

Use symmetry about $y=x$ for inverse-curve areas; subtract the triangle with base and height $b$ from $I$ before doubling. Figure has point $(b, b)$, open endpoint $2a$, and asymptotes $y=a$, $x=a$.

Answer key

(a) No; $R_g\nsubseteq D_f$; $gf:x\mapsto(x^2-6x+8)/(x^2-6x+11)$, $x<4$; $R_{gf}=[-1/2, 1)$. (b) inverse $-\ln(x/a-1)$; printed domain $(a, 2a)$; area $2I-b^2$.

Native diagrams