Question

Let A = {x &epsis; R | −1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by f(x) = x² and g(x) = sin (π x/2). Show that g⁻¹ exists but f⁻¹ does not exist. Also, find g−1.

Answer 1

f is not one-one because
$$\begin{gathered} f\left(-1\right)={\left(-1\right)}^2=1 \\ \mathrm{and}f\left(1\right)=1^2=1 \end{gathered}$$
$\Rightarrow$ -1 and 1 have the same image under f.
$\Rightarrow$ f is not a bijection.
So, f ^-1 does not exist.

Injectivity of g:
Let x and y be any two elements in the domain (A), such that
$$\begin{gathered} g\left(x\right)=g\left(y\right) \\ \Rightarrow \sin \left(\frac{\mathrm{\pi x}}{2}\right)=\sin \left(\frac{\mathrm{\pi y}}{2}\right) \\ \Rightarrow \left(\frac{\mathrm{\pi x}}{2}\right)=\left(\frac{\mathrm{\pi y}}{2}\right) \\ \Rightarrow x=y \end{gathered}$$
So, g is one-one.

Surjectivity of g:
$$\begin{gathered} \text{Range of}g\text{=}\left[\sin \left(\frac{\mathrm{\pi }\left(-1\right)}{2}\right),\sin \left(\frac{\mathrm{\pi }\left(1\right)}{2}\right)\right] \\ \text{=}\left[\sin \left(\frac{-\mathrm{\pi }}{2}\right),\sin \left(\frac{\mathrm{\pi }}{2}\right)\right]\text{=}\left[-1,1\right]\text{=}\mathit{\text{A}}\text{(co-domain of}\mathit{\text{g}}\text{)} \end{gathered}$$
$\Rightarrow$g is onto.
$\Rightarrow$g is a bijection.
So, g^-1 exists.

Also,
$$\begin{gathered} \text{let}{g}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow g\left(y\right)=x \\ \Rightarrow \sin \left(\frac{\mathrm{\pi y}}{2}\right)=x \\ \Rightarrow \left(\frac{\mathrm{\pi y}}{2}\right)={\sin }^{-1}x \\ \Rightarrow y=\frac{2}{\mathrm{\pi }}{\sin }^{-1}x \\ \Rightarrow g^{-1}\left(x\right)=\frac{2}{\mathrm{\pi }}{\sin }^{-1}x[\text{from}\left(1\right)\text{]} \end{gathered}$$