Matching rules are not equal without matching domains · source reference

Printed question

The function $f$ is defined by $f:x\mapsto e^{|2-x|}$, $x\in\mathbb R$, $x>a$, where $a$ is a constant. It is given that $a$ is the least possible value such that $f^{-1}$ exists.

(i) State the value of $a$. [1]

(ii) Hence find $f^{-1}(x)$ and state the domain of $f^{-1}$. [4]

(iii) Sketch on the same diagram the graphs of $y=f(x)$, $y=f^{-1}(x)$ and $y=ff^{-1}(x)$, giving the coordinates of all end points. [3]

It is now given that $a$ is replaced by a constant such that the functions $ff^{-1}$ and $f^{-1}f$ are equal.

(iv) Explain why the new value of $a$ satisfies the equation $a=f(a)$, and find $a$. [2]

(v) The function $g$ is defined by $g:x\mapsto e^{|2-x|}$, $x\in\mathbb Z$. Explain why composite $g^2$ does not exist. [1]

[2021/EJC/Promo/8]

Printed working

(i) $a=2$. The unrestricted exponential absolute-value graph has minimum $(2, 1)$ and intercept $(0, e^2)$.

(ii) Let $y=e^{|2-x|}$. $x>a$, i. e. $x>2\Rightarrow|2-x|=x-2$. Thus $y=e^{x-2}$, $\ln y=x-2$, $x=\ln y+2$. Therefore $f^{-1}(x)=\ln x+2$, $D_{f^{-1}}=R_f=(1, \infty)$.

(iii) Native graphs have open endpoints $f:(2, 1)$, $f^{-1}:(1, 2)$ and $ff^{-1}:(1, 1)$; $ff^{-1}(x)=x$ for $x>1$. All three meet at the fixed point.

(iv) The source writes “As $f^{-1}$ exists, $a>2$”; existence alone allows $a=2$ as well, but the new equal-domain solution is indeed greater than $2$.

$ff^{-1}(x)=f^{-1}f(x)=x$ (same rule). To be equal, also need $D_{ff^{-1}}=D_{f^{-1}f}$ (same domain) ⇒ $D_{f^{-1}}=D_f$ ⇒ $R_f=D_f$ ⇒ $(f(a), \infty)=(a, \infty)$ ⇒ $a=f(a)$. Here $f(a)$ denotes the formula’s limiting endpoint value although $a$ is excluded from the domain. GC gives $a=3.15$ (3 s. f.).

(v) If $g^2$ exists, $R_g\subseteq D_g$. But $g(0)=e^2\in R_g$, and $e^2$ is not an integer, so $e^2\notin D_g$. Hence $R_g\nsubseteq D_g$, and $g^2$ does not exist.

Teacher notes

For $f^{-1}$ to exist, $f$ has to be one-one.

$ff^{-1}(x)=x$; $D_{ff^{-1}}=D_{f^{-1}}=R_f=(1, \infty)$. The three graphs intersect at the same point.

Answer key

(i) 2. (ii) $f^{-1}(x)=\ln x+2$, $D_{f^{-1}}=R_f=(1, \infty)$. (iii) Native three-curve sketch. (iv) $a=3.15$ (3 s. f.). (v) $R_g\nsubseteq D_g$.

Native diagrams