Let $f(x) = (- x^3 + 15)/7$, let $\alpha$ be the number of real roots of $f(x) = f^{-1} (x)$, and let $\beta$ be the number of real roots of $f(x) = x$. What is $|\alpha - \beta|?$
On a quick glance and drawing the graph, the answer seems to be $2$ ($\alpha=3$, $\beta=1$), but the actual answer is $5$ ($\alpha=5$, $\beta=1$). I am unable to find a way to find $\alpha$; please give a detailed solution.
Note: I assumed that finding the roots is not what the OP is after, so I did not detail how can the root values be obtained. I focused on finding the value of $|\alpha - \beta|$. I used Desmos to find the roots.
$$f(x)= \frac{1}{7}\left(-x^{3}+15\right)$$
Since (see notes above by Thomas Andrews and Coffeemath): $f(x)=f^{-1}(x)$ Is equivalent to:
$$ f(f(x))=x$$
And since $\alpha$ is the number of roots for: $f(x)=f^{-1}(x)$, $\alpha$ is the number of roots for:
$$f(f(x))-x=0$$
Which can be written as: $$g_{1}(x)=\left(-\frac{\left(\left(-\frac{x^{3}}{7}+\frac{15}{7}\right)\right)^{3}}{7}+\frac{15}{7}\right)-x=0$$
The above has the real roots: $x=-1.84, x=1,x= 1.58, x=2$ and $x=3.032$ (note that the above is a polynomial of degree $9$). We can conclude that: $$\alpha = 5$$
It is also given that the number of roots for $f(x)-x$ is $\beta$. The expression can be written as:
$$g_{2}(x)=\frac{1}{7}\left(-x^{3}+15\right)-x$$
The above polynomial has 1 real root of value:$1.58$, so: $$\beta = 1$$
$$ |\alpha - \beta| = |5-1| = 4$$