Finding number of solutions of $f(x)=f^{-1}(x)$ where $f(x)=x^3+x-1$

Question

Let $f:\mathbb R\to\mathbb R, f(x)=x^3+x-1$. Find the number of solutions of the equation $f(x)=f^{-1}(x)$.

$f'(x)=3x^2+1\gt0\implies f(x)$ is an increasing function. From this can we say anything about the nature of $f^{-1}(x)$?

I know that a function and its inverse are mirror images in $y=x$. Not sure how to use that here.

Generally, we find inverse function by writing $x$ in terms of $y$, where $y=f(x)$. Here, $$y=x^3+x-1$$

Not able to obtain $f^{-1}(x)$ from this.

Answer 1

$f(x)=f(x)^{-1}$ has always the same set of solutions as $f(x)=x$.

Explanation:

$x^3+x-1$ is crossing $y$ at $-1$ since $0^3+0-1=-1$.

Notice now that first derivative $3x^2+1>1>0$, except at $0$. So the function is going faster than the linear function except at $0$.

So the shape of the function is that it is going from minus infinity crosses $y$ at $-1$ and keeps off to infinity faster than linear function.

A function $f(x)^{-1}$ is a mirror image over $f(x)$ done across $g(x)=x$ so, because of the shape, the mirror of $f(x)$ and $f(x)$ itself can cross only once.

So you have only one solution.

This is all to say that $f(x)=f(x)^{-1}$ has always the same set of solutions as $f(x)=x$, which gives $x^3=1$ in your case.

Answer 2

Since $f(1)=1$, you have $f^{-1}(1)=1$, and therefore the equation $f(x)=f^{-1}(x)$ has, at least, one solution, which is $1$.

It turns out that it is the only one. You always have $f'(x)\geqslant1$; in fact, $f'(x)>1$, unless $x=0$, in which case $f'(x)=1$. So, $f'\bigl(f(x)\bigr)f'(x)$ is always greater than or equal to $1$. In fact, it is always greater than $1$, because it would be $1$ if and only if $f'\bigl(f(x)\bigr)=f'(x)=1$, but\begin{align}f'(x)=1&\implies x=0\\&\implies f(x)=1\\&\implies f'\bigl(f(x)\bigr)>1.\end{align}So, the equation $f\bigl(f(x)\bigr)=x$ can have no more than one solution, since the map $x\mapsto f\bigl(f(x)\bigr)-x$ is strictly increasing (its derivative is always greater than $0$). And$$f(x)=f^{-1}(x)\implies f\bigl(f(x)\bigr)=x.$$

Answer 3

If $f(x)=f^{-1}(x)$ then $f(f(x))=x$, where \begin{eqnarray*} f(f(x))&=&f(x)^3+f(x)-1\\ &=&(x^3+x-1)^3+(x^3+x-1)-1\\ &=&x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+4x-3. \end{eqnarray*} So you want to find the real roots of $$x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+3x-3=0.$$ The above is equivalent to $$(x^3+x-1)^3+(x^3+x-1)-1=x,$$ which shows that the polynomial is divisible by $x^3-1$. Then a quick check shows that $$x^9+3x^7-3x^6+3x^5-6x^4+5x^3-3x^2+3x-3=(x^3-1)(x^6+3x^4-2x^3+3x^2-3x+3).$$ Now to show that $x=1$ is the unique real solution, it suffices to show that the sextic factor has no real roots. It is not hard to express it as a sum of squares: \begin{eqnarray*} x^6+3x^4-2x^3+3x^2-3x+3&=&(x^3-1)^2+3x^4+3x^2-3x+2\\ &=&(x^3-1)^2+3x^4+3(x-\tfrac12)^2+\tfrac54. \end{eqnarray*}

Answer 4

(Motivated by Alex Peter's solution.)
If $f$ is increasing, then the solutions of $f(x)=f^{-1}(x)$ are exactly the solutions of $f(x)=x.$ Let $f(f(a))=a.$ Let $b=f(a).$ Thus, $f(b)=a.$ If $a\lt b,$ then $f(a)\gt f(b),$ a contradiction, (Similar argument for $a\gt b.$) Therefore, $a=b.$ (The decreasing example $f(x)=x^{-1}$ is a nice contrast.) Thus, in the given question, solve $x^3=1$ (as Alex Peter said).