# 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:

$$f(x)=x^{11}+x^8+x^7+x^4+x^3+x+7$$

• What do you mean by "$x$ is replicated"? What's the relationship between the first and the last expression? – Matti P. Feb 13 at 12:54
• Well, in general it won't work (assuming you mean "what if we have more terms"). For instance, $x^3 - x$ doesn't have an inverse. – Arthur Feb 13 at 12:54
• In general it won't have an inverse as a function. However it will have an inverse as a set mapping. – lightxbulb Feb 13 at 12:58

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: $$7\, \left( 7\,n+11 \right) \left( 7\,n+15 \right) \left( 7\,n+19 \right) \left( 7\,n+23 \right) \left( 7\,n-1 \right) \left( 7\,n+3 \right) a_{{n}}-686\, \left( n+1 \right) \left( 14406\,{n}^{5}+ 156065\,{n}^{4}+646212\,{n}^{3}+1264249\,{n}^{2}+1150758\,n+382635 \right) a_{{n+1}}+9604\, \left( n+1 \right) \left( n+2 \right) \left( 5145\,{n}^{4}+54880\,{n}^{3}+215649\,{n}^{2}+368816\,n+230835 \right) a_{{n+2}}-268912\, \left( n+1 \right) \left( n+2 \right) \left( n+3 \right) \left( 490\,{n}^{3}+4655\,{n}^{2}+14679\,n+15345 \right) a_{{n+3}}+16\, \left( n+1 \right) \left( n+2 \right) \left( n+3 \right) \left( n+4 \right) \left( 12353577\,{n}^{2}+ 90592322\,n+166594680 \right) a_{{n+4}}-32\, \left( 4942122\,n+ 20591599 \right) \left( n+5 \right) \left( n+4 \right) \left( n+3 \right) \left( n+2 \right) \left( n+1 \right) a_{{n+5}}+52734400\, \left( n+1 \right) \left( n+2 \right) \left( n+3 \right) \left( n+ 4 \right) \left( n+5 \right) \left( n+6 \right) a_{{n+6}} =0$$