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 ?
 A: Hint: Look at derivative of the function.
A: In order to show that a function is one-to-one, it is sometimes (a lot of times actually) easier to use the derivative rather than showing it directly. A theorem in calculus says that if $f(x)$ is strictly monotone (strictly increasing or strictly decreasing), then it is one-to-one.  The derivative is $f'(x) = 2e^{2x} + 5x^4$. Note that $f'(x) > 0$ since $2e^{2x} > 0$ and $5x^4 \geq 0$. This actually shows that $f(x)$ is strictly increasing since it is $f'(x) > 0$ means that the function $f$ is always "sloped upwards." Therefore, $f(x)$ is one-to-one.
A: Check your subscripts.  Then ask when is the left-hand side positive?  When is the right-hand side positive?
For the second part, remember that $$(f^{-1})\,'(x)=\frac{1}{f\,'(f^{-1}(x))}$$
Just solve the equation $f(y)=2$, and the answer is the reciprocal of the derivative at that point.
A: Hint: Check the derivative has only one sign.
A: You can also prove one-one  this way: (Scroll the  mouse over the covered region below)

  Use the fact : $x_1 \neq x_2 \implies f(x_1) \neq f(x_2)$

A: $f'(x)=2e^{2x}+4x^4>0\,\,$ and 
so function is Ascending(increasing) 
then must be $1-1$ function
A: since derivative of given function is
$f'(x)=2e^{2x} + 5x^4$.ie
$f'(x)>0$ for all x.
so $f(x)$ strictly increasing for all x.
that implies according to mean value theorem
$\frac{f(b)-f(a)}{b-a}$>0.
that implies $f(b)-f(a)$ cannot be zero unless $a=b$
hence proved
