inverse of the function whose exponential value is different $f(x)=x^2+2 $

$f^{-1}(x)= \sqrt {x-2} $ 

$g(x)=x^7+x^3+2$ 

$g^{-1}(x)= ?$

and for more $x^y$?

for example:

$f(x)=x^{11}+x^8+x^7+x^4+x^3+x+7$
In general, you won't have "closed-form" expressions for $f^{-1}$ when $f$ is a polynomial of degree  greater than $4$. You could have series expansions in appropriate intervals.  For example, the series for your first $f^{-1}(y)$ in powers of $y$ is

$$ -1+{\frac{1}{10}}y+{\frac{3}{125}}{y}^{2}+{\frac{99}{12500}}{y}^{3}+{
\frac{1471}{500000}}{y}^{4}+{\frac{145263}{125000000}}{y}^{5}+{\frac{
2978353}{6250000000}}{y}^{6}+{\frac{62619501}{312500000000}}{y}^{7}+{
\frac{2681146413}{31250000000000}}{y}^{8}+{\frac{11637931909}{
312500000000000}}{y}^{9} + \ldots = \sum_{n=0}^\infty a_n y^n$$
where the coefficients $a_n$ satisfy a rather complicated linear recurrence:
$$ 7\, \left( 7\,n+11 \right)  \left( 7\,n+15 \right)  \left( 7\,n+19
 \right)  \left( 7\,n+23 \right)  \left( 7\,n-1 \right)  \left( 7\,n+3
 \right) a_{{n}}-686\, \left( n+1 \right)  \left( 14406\,{n}^{5}+
156065\,{n}^{4}+646212\,{n}^{3}+1264249\,{n}^{2}+1150758\,n+382635
 \right) a_{{n+1}}+9604\, \left( n+1 \right)  \left( n+2 \right) 
 \left( 5145\,{n}^{4}+54880\,{n}^{3}+215649\,{n}^{2}+368816\,n+230835
 \right) a_{{n+2}}-268912\, \left( n+1 \right)  \left( n+2 \right) 
 \left( n+3 \right)  \left( 490\,{n}^{3}+4655\,{n}^{2}+14679\,n+15345
 \right) a_{{n+3}}+16\, \left( n+1 \right)  \left( n+2 \right) 
 \left( n+3 \right)  \left( n+4 \right)  \left( 12353577\,{n}^{2}+
90592322\,n+166594680 \right) a_{{n+4}}-32\, \left( 4942122\,n+
20591599 \right)  \left( n+5 \right)  \left( n+4 \right)  \left( n+3
 \right)  \left( n+2 \right)  \left( n+1 \right) a_{{n+5}}+52734400\,
 \left( n+1 \right)  \left( n+2 \right)  \left( n+3 \right)  \left( n+
4 \right)  \left( n+5 \right)  \left( n+6 \right) a_{{n+6}}
=0$$
In general, you won't have "closed-form" expressions for $f^{-1}$ when $f$ is a polynomial of degree  greater than $4$. You could have series expansions in appropriate intervals.  For example, the series for your first $f^{-1}(y)$ in powers of $y$ is

$$ -1+{\frac{1}{10}}y+{\frac{3}{125}}{y}^{2}+{\frac{99}{12500}}{y}^{3}+{
\frac{1471}{500000}}{y}^{4}+{\frac{145263}{125000000}}{y}^{5}+{\frac{
2978353}{6250000000}}{y}^{6}+{\frac{62619501}{312500000000}}{y}^{7}+{
\frac{2681146413}{31250000000000}}{y}^{8}+{\frac{11637931909}{
312500000000000}}{y}^{9} + \ldots = \sum_{n=0}^\infty a_n y^n$$
where the coefficients $a_n$ satisfy a rather complicated linear recurrence:
$$ 7\, \left( 7\,n+11 \right)  \left( 7\,n+15 \right)  \left( 7\,n+19
 \right)  \left( 7\,n+23 \right)  \left( 7\,n-1 \right)  \left( 7\,n+3
 \right) a_{{n}}-686\, \left( n+1 \right)  \left( 14406\,{n}^{5}+
156065\,{n}^{4}+646212\,{n}^{3}+1264249\,{n}^{2}+1150758\,n+382635
 \right) a_{{n+1}}+9604\, \left( n+1 \right)  \left( n+2 \right) 
 \left( 5145\,{n}^{4}+54880\,{n}^{3}+215649\,{n}^{2}+368816\,n+230835
 \right) a_{{n+2}}-268912\, \left( n+1 \right)  \left( n+2 \right) 
 \left( n+3 \right)  \left( 490\,{n}^{3}+4655\,{n}^{2}+14679\,n+15345
 \right) a_{{n+3}}+16\, \left( n+1 \right)  \left( n+2 \right) 
 \left( n+3 \right)  \left( n+4 \right)  \left( 12353577\,{n}^{2}+
90592322\,n+166594680 \right) a_{{n+4}}-32\, \left( 4942122\,n+
20591599 \right)  \left( n+5 \right)  \left( n+4 \right)  \left( n+3
 \right)  \left( n+2 \right)  \left( n+1 \right) a_{{n+5}}+52734400\,
 \left( n+1 \right)  \left( n+2 \right)  \left( n+3 \right)  \left( n+
4 \right)  \left( n+5 \right)  \left( n+6 \right) a_{{n+6}}
=0$$