Self-inverse functions alternate by parity · source reference

Printed question

A function $f$ is said to be a self-inverse function if $f(x)=f^{-1}(x)$. The function $f$ is defined by $f(x)=\frac{ax+b}{cx-2}$, $x\in\mathbb R$, $x\ne2/c$, where $a, b, c, d$ are non-zero constants.

(i) By finding $f^{-1}(x)$, show that $a=2$ for $f$ to be self-inverse. [2]

For the rest use $a=2, b=3, c=5$. (ii) Find $f^2(x)$. [1] (iii) Evaluate $f^{71}(4)$. [2]

The function $g$ is defined by $g(x)=2x^2-3$, $x\in\mathbb R$. (iv) Explain why composite $fg$ does not exist. [2] (v) Find exact solutions of $gf(x)=5$. [3]

[2020/CJC/Promo/10]

Printed working

(i) $y=(ax+b)/(cx-2)$ ⇒ $cxy-2y=ax+b$ ⇒ $x(cy-a)=b+2y$ ⇒ $x=(b+2y)/(cy-a)=(2y+b)/(cy-a)$.

Hence $f^{-1}(x)=(2x+b)/(cx-a)$. Comparing gives $a=2$ (under the inverse-existence/nonconstant premise: the determinant $-2a-bc$ must be nonzero).

(ii) $f=f^{-1}$ ⇒ $ff=ff^{-1}$ ⇒ $f^2(x)=ff^{-1}(x)=x$, for $x\ne2/5$.

(iii) $f^{71}(x)=f^{35\times2+1}(x)=f(x)$; $f^{71}(4)=f(4)=[2(4)+3]/[5(4)-2]=11/18$.

(iv) For $fg$ to exist, $R_g\subseteq D_f$. $R_g=[-3, \infty)$, $D_f=(-\infty, 2/5)\cup(2/5, \infty)$. Since $R_g\nsubseteq D_f$, $fg$ does not exist.

(v) $gf(x)=5$ ⇒ $g((2x+3)/(5x-2))=5$ ⇒ $2((2x+3)/(5x-2))^2-3=5$ ⇒ $((2x+3)/(5x-2))^2=4$.

Method 1: $2x+3=\pm2(5x-2)$. Positive sign gives $2x+3=2(5x-2)\Rightarrow8x=7\Rightarrow x=7/8$. Negative gives $2x+3=-2(5x-2)\Rightarrow12x=1\Rightarrow x=1/12$.

Method 2: square relation gives $4x^2+12x+9=4(25x^2-20x+4)$ ⇒ $96x^2-92x+7=0$.

$$x=\frac{92\pm\sqrt{8464-4(96)(7)}}{2(96)}=\frac{92\pm76}{192}=\frac78\text{ or }\frac1{12}.$$

Both avoid the excluded $2/5$. The printed unused constant $d$ in the statement is retained; no $d$ is present in the formula.

Teacher notes

Compare $f(x)=f^{-1}(x)$: $\frac{ax+b}{cx-2}=\frac{2x+b}{cx-a}\Rightarrow a=2$.

Iterate in pairs: $f^{71}=f(f^2)(f^2)\cdots(f^2)$ with 35 pairs; removing one identity pair leaves 34 pairs, and continuing gives $f$.

The range/domain number-line diagram has $R_g$ starting at closed $-3$, while $D_f$ excludes the open hole $2/5$.

Answer key

(i) $f^{-1}(x)=(2x+b)/(cx-a)$. (ii) $f^2(x)=x$. (iii) $11/18$. (iv) $R_g\nsubseteq D_f$. (v) $7/8$ or $1/12$.

Native diagrams