Composition conditions and a hidden constant case · source reference

Printed question

Functions $f:x\mapsto(x^2-6x-6)/(x+1)$, $x\in\mathbb R$, $x\ne-1$; $g:x\mapsto(ax+1)/(x+b)$, $x\in\mathbb R$, $x\ne-b$, where $a, b$ are constants.

(i) Sketch $y=f(x)$, giving coordinates of turning points and equations of asymptotes. Write down the range of $f$. [3]

(ii) Find the value of $a$ and range of $b$ such that both $fg$ and $gf$ exist. [4]

(iii) Find $g^{-1}(x)$. Given $g^{-1}(x)=g(x)$ for all real $x$, $x\ne-b$, find $b$ in terms of $a$. [3]

[2019/TJC/Promo/9]

Printed working

(i) $f(x)=x-7+1/(x+1)$. Native source graph has turning points $(0, -6)$ and $(-2, -10)$, vertical asymptote $x=-1$, oblique $y=x-7$. Range $(-\infty, -10]\cup[-6, \infty)$.

(ii) Printed nonconstant argument:

$$g(x)=\frac{ax+1}{x+b}=a+\frac{1-ab}{x+b}.$$

$R_g=\mathbb R\setminus\{a\}$, $D_f=\mathbb R\setminus\{-1\}$. For $fg$ to exist, $R_g\subseteq D_f$, i. e.$\mathbb R\setminus\{a\}\subseteq\mathbb R\setminus\{-1\}$, hence $a=-1$.

$R_f=(-\infty, -10]\cup[-6, \infty)$, $D_g=\mathbb R\setminus\{-b\}$. For $gf$ to exist, $R_f\subseteq D_g$ ⇒$-10<-b<-6$ ⇒$6<b<10$.

Constant-family qualification

The displayed range $\mathbb R\setminus\{a\}$ requires $1-ab\ne0$. Literal (ii) also permits $ab=1$: then $g(x)=a=1/b$ on its punctured domain, $R_g=\{1/b\}$. Both composites exist for $6<b<10$, since $1/b\ne-1$ and $-b$ is outside $R_f$. Part (iii) requires an inverse and therefore excludes constant $g$. If that premise applies throughout, the printed nonconstant answer $a=-1$ is the intended answer.

(iii) Let $y=(ax+1)/(x+b)$ ⇒$xy+by=ax+1$ ⇒$x=(1-by)/(y-a)$ ⇒$g^{-1}(x)=(1-bx)/(x-a)$, $x\ne a$, provided $1-ab\ne0$.

Given $g^{-1}=g$, $(1-bx)/(x-a)=(ax+1)/(x+b)$. By comparison, $b=-a$. With $b=-a$ the determinant is nonzero($1+a^2>0$), so the inverse exists and domains agree.

The short key sketch labels oblique line“$y=x$”, whereas the detailed working labels $y=x-7$; the latter is consistent with $f(x)=x-7+1/(x+1)$. Both printed labels are recorded, and native correct curve uses $y=x-7$.

Teacher notes

No separate teacher sidebar is printed. The rational graph for $g$ illustrates nonconstant case with asymptotes $x=-b$, $y=a$; the constant case has a removable hole at $x=-b$ instead of a vertical asymptote.

Answer key

(i) $(-\infty, -10]\cup[-6, \infty)$, turning points and asymptotes in the native figure. Shortkey oblique label“$y=x$” is corrected separately above. (ii) Printed $a=-1$, $6<b<10$, constant-family qualification above. (iii) $b=-a$.

Native diagrams