Assume that $f(x)=x$ has no real roots where

$$f(x) = ax^2+bx+c$$ Prove that $f(f(x))=x$ has no real roots as well.

What I've done is, calculating $f(f(x))$:


and putting $\Delta=b^2-4ac<0$ which seems quite time consuming. Is that the right thing to do?

  • $\begingroup$ Isn't the fact that $f(x)$ isn't real or $b^2-4ac < 0$ enough for proving $f(f(x))$ can't be real? As the graph of $f(x)$ won't ever touch the x-axis. Please do correct me, If I'm wrong. $\endgroup$ – Ishaan Singh May 1 '12 at 21:05
  • $\begingroup$ @IshaanSingh: That's what I am asking. I didn't come up with anything doing that. $\endgroup$ – Gigili May 1 '12 at 21:07
  • $\begingroup$ Suppose that $f(f(x)) = x$ for some point $x$, what is the possible values for $f(x)$? Since $f(f(x)) = a(f(x))^2 + bf(x) + c = x$, this means $$ f(x) = \frac{-b \pm \sqrt{b^2 - 4a(c-x)}}{2a} = a x^2 + bx + c $$ which, by putting things together, means that $$ -b \pm \sqrt{b^2 - 4a(c-x)} = 2a^2 x^2 + 2abx + c, \quad \Longrightarrow \quad b^2 - 4a(c-x) =(2a^2 x^2 + 2abx + c-b)^2. $$ Perhaps a solution could come out of this? I don't have time to continue this approach though. Maybe later. $\endgroup$ – Patrick Da Silva May 1 '12 at 21:12
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    $\begingroup$ @PatrickDaSilva: That's what I have done and no solution came out of it! $\endgroup$ – Gigili May 1 '12 at 21:15
  • $\begingroup$ @IshaanSingh: It is. I think it's pretty obvious. $\endgroup$ – Gigili May 1 '12 at 21:19

Write $g(x)=f(x)-x$. Then $g$ is continuous and never zero, so it must be either always positive or always negative.

Now $f(f(x))-x = g(f(x))+g(x)$, which is always positive if $g$ is always positive, or always negative if $g$ is always negative. In either case, it's never zero, so $f(f(x))$ is never equal $x$.



$f(x) - x$ is always of the same sign...

To use the hint:

If $f(x) \gt x$ for all $x$, then $f(f(x)) \gt f(x) \gt x$.

Similarly, if $f(x) \lt x$ we can show that $f(f(x)) \lt x$.

(For more details, see Arturo's answer).

Since this is tagged algebra precalculus, here is a continuity free proof to show that $f(x) - x$ is of the same sign.

We will first show that, if $g(x) = px^2 + qx + r$, $p \gt 0$, is a quadratic and if there is a real number $s$ such that $g(s) \lt 0$ then $g(x) = 0$ has a real root.

By completing the square, we have

$$g(s) = p\left(s + \frac{q}{2p}\right)^2 + r - \frac{q^2}{4p} \lt 0$$ i.e. $$0 \le p\left(s + \frac{q}{2p}\right)^2 \lt -r + \frac{q^2}{4p}$$

Thus $q^2 \gt 4pr$ and $g(x) = 0$ has a real root.

The case $p=0$ is easy to deal with.

Now let $s,t$ be such that $f(s) \lt s$ and $f(t) \gt t$. If $a \ge 0$, we can apply the above to $g(x) = f(x) - x$, else, we apply it to $g(x) = x - f(x)$

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    $\begingroup$ So, $f$ being continuous is enough. (Hope I haven't missed something). $\endgroup$ – Aryabhata May 1 '12 at 21:11

Note that $y=f(x)$ never crosses the line $y=x$ because $f$ is continuous. That means that $f(a)$ is always greater than $a$, or $f(a)$ is always smaller than $a$.

If $f(a)\gt a$ for all $a$, then $f(f(x))\gt f(x)\gt x$ for all $x$. If $f(a)\lt a$ for all $a$, then $f(f(x)) \lt f(x) \lt x$ for all $x$. In particular, you never have $f(f(x))=x$.

In fact, for all natural numbers $n$, $f^{\circ n}(x)=x$ has no solutions, for the same reason.


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