number of solutions of $f(f(f(f(x))))$

Hint : $$ f(x) = x^2+10x+20 = (x+5)^2 - 5 $$

so $$ f(f(x)) = ((x+5)^2 - 5 + 5)^2 - 5 = (x+5)^4 - 5 $$ $$f(f(f(x))) = (x+5)^8 - 5 $$ ... and it should be easier to analyse $f^{(4)} (x)$ now.


The minimum of $f(x)$, a concave up quadratic, occurs at $x = -\frac{-10}{2(1)} = -5$, and $f(-5) = -5$. Then the turning point of $f(f(x))$ will occur at $f(f(-5)) = f(-5) = -5$, and so on until $f^4 (x)$ (which denotes $f(f(f(f(x))))$. There is only one minimum of $f(x)$, so there is only one turning point of $f^4 (x)$.

$f^4 (x)$ is also concave up, since the leading term of $f(x), f(f(x)) \cdots$ does not change sign, so the leading term is positive.

Now of a concave up function with only one turning point, how many roots are there in total?


An alternative method to those already presented:

Consider first a simpler case: when is $f^2(x)=0$? (I let $f^2(x)$ denote $f(f(x))$, or, equivalently, $(f \circ f)(x)$.)

It will be whenever $f(x) \in f^{-1}(0)$. (Recall that $f^{-1}(a) = \{ x \mid f(x)=a \}$, i.e. this is a set.)

A similar idea applies to your case. $f^4(x) = 0$ if and only if $f(x) \in (f^{-1})^3(0)$. This is something we can proceed through iteratively.

  • First, what elements are in $f^{-1}(0)$? That is, which elements does $f$ map to $0$? As you have determined, these are $-5 \pm \sqrt 5$.

  • Next, what elements are in $(f^{-1})^2(0) = f^{-1}(-5 \pm \sqrt 5)$? This means that $$x^2 + 10x + 20 = -5 \pm \sqrt 5$$ so just solve for $x$. You get $-5 \pm \sqrt[4]5$ for the positive root, and $-5 \pm i \sqrt[4]5$ for the negative root. Only the former is relevant.

  • Iterate one more time. Then $$x^2 + 10x + 20 = -5 \pm \sqrt[4]5$$ and we solve for $x$. The negative solution gives us more complex numbers, but the positive gives us $-5 \pm \sqrt[8]5$.

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Functions

Contest Math

Quadratics

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