If f(x) is an invertible function and $g(x)=2f(x)+5$, then the value of ${{g}^{-1}}(x)$ is
A. $2{{f}^{-1}}(x)-5$
B. $\dfrac{1}{2{{f}^{-1}}(x)+5}$
C. $\dfrac{1}{2}{{f}^{-1}}(x)+5$
D. ${{f}^{-1}}\left( \dfrac{x-5}{2} \right)$

Hint: When an invertible function is inverted, the domain and range swap places. Here we have given an invertible function and $g(x)=2f(x)+5$ in order to obtain the value of ${{g}^{-1}}(x)$ replace $x$ by ${{g}^{-1}}(x)$. Use the given functions, and replace the variable $x$ that is present in the external function with the internal function. Finally, simplify the function that was obtained.

Formula Used: Let $f$ and $g$ represent two different functions.
If $f(g(x)) = x$ and $g(f(x)) = x$, then
Therefore, $g$ is the inverse of $f$, and $f$ is the inverse of $g$.

Complete step by step solution: We have given that $f(x)$ is an invertible function and $g(x)=2f(x)+5$.
Replace $x$ by ${{g}^{-1}}(x)$ we get;
$\Rightarrow \mathrm{gg}^{-1}(\mathrm{x})=2 \mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right)+5 \\
\Rightarrow \mathrm{x}=2 \mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right)+5 \\
\Rightarrow \mathrm{x}-5=2 \mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right) \\
\Rightarrow \mathrm{f}\left(\mathrm{g}^{-1}(\mathrm{x})\right)=\dfrac{\mathrm{x}-5}{2} \\
\Rightarrow \mathrm{g}^{-1}(\mathrm{x})=\mathrm{f}^{-1}\left(\dfrac{\mathrm{x}-5}{2}\right)$
Hence, the value of ${{g}^{-1}}(x) =\mathrm{f}^{-1}\left(\dfrac{\mathrm{x}-5}{2}\right)$.

So, Option ‘D’ is correct

Note: Our function is invertible if it is both One to One and Onto, and it is not if either of those conditions holds. Remember that each statement for the composition uses the same letters in the same order.
You should begin with the function $g$ since $f (g(x))$ makes this clear (innermost parentheses are done first).
It should be noticed that if the range of $f$ is a subset of $g$, then $gof$ exists. In the same way, if the range of $g$ is a subset of domain $f$, then $fog$ exists.