Prove $f(x)= e^{2x} + x^5 + 1$ is one to one Prove $$f(x)= e^{2x}   + x^5 + 1$$ is one to one. 
So my solution is: Suppose $$ f(x_1)=f(x_2),$$
then I am stuck here: $$e^{2x_1}-e^{2x_2}=x^5_1 -x^5_2.$$  How do I proceed?
Also after that I found out that $$(f^-1)'(2)= 4.06$$ but how do I find $$(f^-1)''(2)$$ Differentiatie $(f^-1)'(2)$ again ?

$f'(x)=2e^{2x}+4x^4>0\,\,$ and 
so function is Ascending(increasing) 
then must be $1-1$ function
