How to find the inverse function for $y = x^5 +2x + 1$ I'm using a self study pre-calculus book and have to invert function $y = x^5 + 2x + 1$. I know I have to swap the variables and put everything on one side of the equation such as $y^5 + 2y -x + 1 = 0$ but from here I don't know what to do.  All of the examples I can find on the internet either use calculus which I don't know or show examples with the degree of $2$ rather than $5$ and so they use the quadratic equation to solve for $y$.  So I am stuck.
 A: You either misread the book, misunderstood the bigger picture or it's playing a cruel joke on you because there's no closed form inverse of this function in terms of elementary functions.  The form that exists and can be found using a CAS (in terms of the generalized hypergeometric function) certainly could not be found by a student at that level.
A: I'm thinking you could solve this by making use of the geometric interpretation of the inverse of a function, namely that its graph is symmetric with respect to a line at $45^{\circ}$.
In principle, I think you could compute this symmetric function by rotating each point such that it lands at the corresponding position. Take the image below as an example. Given the graph of the $y=x^5+2x+1$ function, you can take a random point and notice that you could rotate it clockwise by $2\alpha$ to get to the point corresponding to the inverse graph. This could be done, at least I think it could, at a pre-calculus level, but you kind of have to derive the formula for the rotation and then generalize it for all the points, except the one where the graph intersects the line at $45^{\circ}$.

One way of doing it, but not at a pre-calculus level is by using the rotation matrix
$$R(-2\alpha) = \left(\matrix{\cos(2\alpha) & \sin(2\alpha) \\ -\sin(2\alpha) & \cos(2\alpha)}\right)$$
which can be applied as follows
$$\left(\matrix{\cos(2\alpha) & \sin(2\alpha) \\ -\sin(2\alpha) & \cos(2\alpha)}\right) \cdot \left(\matrix{x \\ x^5+2x+1}\right)=\left(\matrix{x' \\ y'}\right)$$
with $\alpha = \pi/4-\arctan \left(\frac{x}{x^5+2x+1}\right)$.
If I did not make some errors in my reasoning, this should at least give you the coordinates for the inverse and then you could in principle consider the job done.
However, I repeat, I am not sure you are supposed to have to derive the formula for the rotation, even more so to use it as I have done above using a vector-matrix approach. Still, one could argue that it is just geometry and some trigonometry so ... pre-calculus?
