Consider $f:{{\mathbf{R}}^{+}}\to \left[ 4,\infty \right]$ given by $f(x)={{x}^{2}}+4$ : Show that $f$ is invertible with the inverse ${{f}^{-1}}$ of $f$ given by ${{f}^{-1}}(y)=\sqrt{y-4}$ where $\mathbf{R}^+$ is the set of all non-negative real numbers.

Hint: Use the definition of an invertible function to identify the conditions you must show (i.e. $f$ is one-one, onto function). Afterwards, let $f(x)=y$ and rewrite $x$ in terms of a function of $y$ to obtain the inverse ${{f}^{-1}}$ of $f$ .

Complete step by step answer:
We observe that if ${{x}_{1}},{{x}_{2}}\in {{\mathbf{R}}^{+}}$ are arbitrary elements such that ${{x}_{1}}={{x}_{2}}$ then $x_{1}^{2}+4=x_{2}^{2}+4\Rightarrow f({{x}_{1}})=f({{x}_{2}})$ . Hence $f$ is a well-defined function.
Now, we’ll show that $f$ is one-one. For some ${{x}_{1}},{{x}_{2}}\in {{\mathbf{R}}^{+}}$ such that $f({{x}_{1}})=f({{x}_{2}})$ , then
$x_{1}^{2}+4=x_{2}^{2}+4\Rightarrow x_{1}^{2}=x_{2}^{2}\Rightarrow {{x}_{1}}=\pm {{x}_{2}}$
However, since ${{x}_{1}},{{x}_{2}}\in {{\mathbf{R}}^{+}}$ i.e. the set of all non-negative real numbers, we see that ${{x}_{1}}=-{{x}_{2}}$ (a negative real number) is not possible. Therefore ${{x}_{1}}$ must be equal to ${{x}_{2}}$.
In conclusion, we have shown that $f({{x}_{1}})=f({{x}_{2}})\Rightarrow {{x}_{1}}={{x}_{2}}$i.e. $f$ is one-one function.
Now, we’ll show that $f$ is onto.
Let $y=f(x)$ for some $x\in {{\mathbf{R}}^{+}}$ . Then $y={{x}^{2}}+4\Rightarrow {{x}^{2}}=y-4\Rightarrow x=\sqrt{y-4}$ (we ignored the other possibility since $x$ cannot be negative).
Clearly, for all $y\in \left[ 4,\infty \right]$, there exists $x\in {{\mathbf{R}}^{+}}$ such that $f(x)=f\left( \sqrt{y-4} \right)={{\left( \sqrt{y-4} \right)}^{2}}+4=y-4+4=y$
Hence $f$ is onto function.
Since $f$ is one-one, onto function, we conclude that ${{f}^{-1}}$ exists.
To find ${{f}^{-1}}$ , we see that $y=f(x)\Rightarrow x={{f}^{-1}}(y)$ . However, we established that $y=f(x)\Rightarrow y={{x}^{2}}+4\Rightarrow x=\sqrt{y-4}$ .

Comparing these equations, we conclude that $x={{f}^{-1}}(y)=\sqrt{y-4}$ .

Note: It is important to follow the specified procedure in this solution in order to show that a function is one-one, onto and hence, is invertible. This is necessary to ensure that you receive full credit in the examination. Afterwards, we simply use the expression obtained while showing the function is onto to find the inverse of the associated function.