Question

Consider f : R → R₊ → [4, ∞) given by f(x) = x² + 4. Show that f is invertible with inverse f⁻¹ of f given by f⁻¹ $\left(x\right)=\sqrt{x-4}$, where R⁺ is the set of all non-negative real numbers.

Answer 1

Injectivity of f :
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)
$$\begin{gathered} \Rightarrow x^2+4=y^2+4 \\ \Rightarrow x^2=y^2 \\ \Rightarrow x=y\left(\text{as co-domain as}{R}^{+}\right) \end{gathered}$$
So, f is one-one.

Surjectivity of f :
Let y be in the co-domain (Q), such that f(x) = y

$$\begin{gathered} \Rightarrow x^2+4=y \\ \Rightarrow x^2=y-4 \\ \Rightarrow x=\sqrt{y-4}\in R \end{gathered}$$

$\Rightarrow$f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f ^-1:
$$\begin{gathered} \text{Let}{f}^{-1}\left(x\right)=y...\left(1\right) \\ \Rightarrow x=f\left(y\right) \\ \Rightarrow x=y^2+4 \\ \Rightarrow x-4=y^2 \\ \Rightarrow y=\sqrt{x-4} \\ \text{So,}{f}^{-1}\left(x\right)=\sqrt{x-4}[\text{from}\left(1\right)] \end{gathered}$$