A function $\mathrm{f}$ is said to be self-inverse if $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$

Question

A function $\mathrm{f}$ is said to be self-inverse if $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$ for all $x$ in the domain of $\mathrm{f}$. It is given that $\mathrm{g}$ is a self-inverse function and is defined by $\mathrm{g} : x \mapsto \frac{x+a}{3x+b}$, for $x \in \mathbb{R},\, x > -\frac{b}{3}$, where $a$ and $b$ are constants and $\mathrm{g}(1) = 5$. Show that $a = 9$ and $b = -1$. Find $\mathrm{g}^{2021}(1)$. The function $\mathrm{h}$ is defined by $$\mathrm{h} : x \mapsto \left\vert{} 1-x \right\vert{}(x+5),\text{ for}\hspace{0.5em} x \in \mathbb{R},\, x \ge 2.$$ Given that $\mathrm{h}^{-1}$ exists, find $\mathrm{h}^{-1}$ in similar form. Show that $\mathrm{gh}$ exists and find the exact range of $\mathrm{gh}$.

Accepted Answer

(i) $a=9$, $b=-1$ (ii) $5$ (iii) $\mathrm h^{-1}(x)=-2+\sqrt{x+9}$ for $x\ge7$ (iv) $R_{\mathrm{gh}}=\left(\dfrac13,\dfrac45\right]$

Functions - Modulus Functions

Functions - Modulus Functions

Video Solution:

Solution:

Solution locked