The Inverse of $y=x^5-x^3+x$ in terms $y=$
I tried finding the inverse of $y=x^5-x^3+x$ in terms of getting there to be only one y value. Such that y equals some function of x. I know that the original function is a one to one function because it passed the horizontal line test. Thus it must have a inverse function.
The first thing I did was switch the y's and x's $$x=y^5-y^3+y$$ The next thing I did was take the derivative of the function because I thought I could maybe try to take the integral after which might help me. $$\frac{dy}{dx}(x=y^5-y^3+y)$$ $$\frac{dy}{dx}=\frac{1}{5y^4-3y^2+1}$$ I'm confused on how to take the integral of the function and that is as far as I got. If taking the integral of the function is not the right way can some please show me what is?
$\endgroup$2 Answers
$\begingroup$We have :
$$f(x) = x^5 - x^3 + x$$
Then, the derivative of $f$ is :
$$f'(x) =5x^4 - 3x^2 + 1$$
It's $f'(x) > 0$ $\forall x \in \mathbb R$ which means that $f(x)$ is strictly increasing which points to $f$ being $1-1$, thus you're correct, the function $f(x)$ has an inverse.
Take into account though, that the existence of an inverse function does not always mean that you can calculate the function within terms of standard mathematical functions.
This is one of these cases, as you cannot solve : $y=x^5 -x^3 + x$ with respect to $x$, so that you can exchange the variables and find $f^{-1}(x)$.
If you want a look at the graph of the inverse, check here!
$\endgroup$ $\begingroup$I haven't checked but $x^5-x^3+x-y$ if probably not solvable by radicals for most values of $y$. So, there is no simple formula for $x$ in terms of $y$.
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