For $x>0$ let $f(x)=x^{2/3}(6-x)^{1/3}$, then find the number of solutions of $f(x)=f^{-1}(x); x\in (0,4)$

Question

For $x>0$ let $f(x)=x^{2/3}(6-x)^{1/3}$, then find the number of solutions of $f(x)=f^{-1}(x); x\in (0,4)$

$\displaystyle f'(x)=\frac{2(6-x)^{1/3}}{3x^{1/3}}+\frac{x^{2/3}}{3(6-x)^{1/3}}$, since $f'(x)$ is always positive this means that $f(x)$ is an increasing function.

Let $f(x)=y$, then $y^3=x^2(6-x)=6x^2-x^3$, is there any way to express $x$ in terms of $y$? How can we proceed after this?

Answer 1

Let us show first that for a strictly increasing function $f$

• (A) $x$ is a solution for the equation $f(x)=f^{-1}(x)$

if and only if

• (B) $x$ is a solution for the equation $x=f(x)$.

One direction is clear. The other one. So let us start with (A) and show (B). First, $f(x)=f^{-1}(x)$ is equivalent with $f(f(x))=x$. (We apply $f$, which is injective.)

Assume (B) is false. (We get soon a contradiction.) Then we have either $x<f(x)$ or $f(x)<x$.

• The first case, $x<f(x)$. Then applying $f$ we get $f(x)<f(f(x))$. Put together, $x<f(x)<f(f(x))$, a contradiction.

• The second case, $f(x)<x$. Then applying $f$ we get $f(f(x))<f(x)$. Put together, $f(f(x))<f(x)<x$, a contradiction.

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It remains now to solve $y^3=x^2(6-x)$ on $(0,4)$, knowing $y=x$, this leads to $x=6-x$, so $x=3$.

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The above solution solves the problem without addressing the intermediate question: "... is there any way to express $x$ in terms of $y$?" - well, i will try to say some words on this, although it may seem completely useless. The following "explains" somehow the relation between $x$ and $y$ by giving a parametrization of the curve $y=f(x)$.

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Observe first that for $x>0$ the corresponding value of $y:=f(x)$ is also $>0$. So we may, and do build the fraction $t=y/x$. So $y=tx$.

(This is the usual trick in algebraic geometry, to try to parametrize a curve, here the algebraic curve with (singular) equation $y^3=x^2(6-x)$, by the slope passing through the (singular) point $(x_0,y_0)=(0,0)$. This may make some sense for some readers, please ignore if this makes no sense. But this allows me to refer to Maria Agnesi, she used and popularized this "trick" in some special examples, she is maybe not the first one, but she is the one i know.)

Then we get successively for $x\in(0,4)$, $y=f(x)>0$, first written down in a formal computation, the result of the computation will be used only: $$ \begin{aligned} y^3 &= x^2(6-x)\ ,\\ t^3x^3 &= x^2(6-x)\ ,\\ t^3x &= 6-x\ ,\\ (t^3+1)x &= 6\ ,\\ x &=\frac 6{t^3+1}\ , \qquad\text{ so}\\ y &= tx= \frac {6t}{t^3+1}\ , \end{aligned} $$ and in this way we get a ((bi)rational) parametrization of the curve.

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In other words, the graph $y=f(x)$ is parametrized (on the interval $x\in(0,4)$) by the points of the shape $$ P(t) := \left(\ \frac 6{t^3+1}\ ,\ \frac {6t}{t^3+1}\ \right)\ . $$ (The parameter $t$ lives in the corresponding interval $(0,2^{-1/3})$ i think...)

A point on the graph of the inverse function, drawn by reflection in the first quadrant bisector, has "switched" coordinates, so it is a point of the shape $$ Q(u) := \left(\ \frac {6u}{u^3+1}\ ,\ \frac {6}{u^3+1}\ \right)\ . $$ Let us now intersect the two graphs. This means, we are searching for suitable parameters $t,u$ (in the appropriate range), so that $P(t)=Q(u)$.

Assume such a condition is satisfied. Then we build the quotient $y/x$, getting, after denominator simplifications $$ \frac {6t}6=\frac6{6u}\ ,$$ so $u=1/t$. Let us plug this relation in $P(t)=Q(u)$, we get $$ \begin{aligned} \left(\ \frac 6{t^3+1}\ ,\ \frac {6t}{t^3+1}\ \right) &=P(t)\\ &=Q(1/t)\\ &= \left(\ \frac {6/t}{1/t^3+1}\ ,\ \frac {6}{1/t^3+1}\ \right) \\ &= \left(\ \frac {6t^2}{1+t^3}\ ,\ \frac {6t^3}{1+t^3}\ \right) \ . \end{aligned} $$ This implies $t=1$. So $x=y=6/2=3$. So getting the point(s) of intersection of the two graphs, means getting the point(s) of intersection with $y=x$, i.e. the fixed points... And there is only one fixed point, $3$, well, we have to check this. (Since all the implications above go in one direction, finally the point $(3,3)$ is the only possible intersection of the graphs, but this has to be checked.) Indeed: $f(3) = 3^{2/3}(6-3)^{1/3} =3^{2/3\ +\ 1/3}=3^1=3$.

(In this case, we intersect two "very particular", indeed reflected curves, so solving algebraically does not make too much sense. But in general...)

Answer 2

As you checked, for $x \in (0,4)$, $f(x)$ is injective. Thus $f^{-1}(x)$ can be defined. Now note that, $f^{-1}(x)$ is merely a reflection of $f(x)$ about the line $y=x$. Hence, if $f(x)=f^{-1}(x)$, then this means that both curves meet, which can only happen on the line $y=x$.

Thus, we can simply look for solutions of $f(x)=x$ in the given interval. Clearly, $x=3$ is the only possible solution.

Answer 3

It is just as well to work with $g(x)=f(6x)/6$ over $(0,2/3)$. So let

$$p(x,y) = y^3 - y^2 + x^3$$ and $$q(x,y) = y^3 + x^3 - x^2$$ so that $y=g(x)$ implies $q(x,y)=0$ and $y=g^{-1}(x)$ implies $p(x,y)=0$. Using generalized Euclidean division to eliminate $y$ gives

$$(y^2-(x-1)y+x^2)p(x,y) - (y^2-xy+x^2+x-1)q(x,y) = r(x)$$ where $$r(x) = x^2(2x-1)$$ so that $g(x)=g^{-1}(x)$ implies $r(x)=0$. Of the roots of $r$, only $x=\tfrac{1}{2}$ is admissible, corresponding to $x=3$ for the unscaled function.