Min of $p(p(p(p(x)))=$ where $p(x)=x^2+6x+7$ If $p(x)=x^2+6x+7$ what is the min of $$p(p(p(p(x)))$$
I get stuck on this . Let me show you what I tried .$$p(x)=(x+3)^2-2 \to \min\{p(x)\}=-2$$
then 
$$p(p(x))=(((x+3)^2-2)+3)^2-2=((x+3)^2+1)^2-2 \\\to \min\{p(p(x))\}=-1$$ I can do this again ,but it comes more complicated ...
  Is there a trick to find $\min\{p(p(p(p(x)))\}$ ?
    Is there a way to find $\min\{\underbrace{p(p(p...(p(x)))}_{n-times}\}$ ?  
 A: The minimum of $p(x)$ is attained at $x=-3$, so the first strategy would be to look for $x$ such that $p(x)=-3$, so that you can feed $p(p(x))$ its minimum argument. But the problem here is that $-3$ is never attained for $p(x)$, since $p(-3)=-2$. What you could do is write $p(x)$ as
$$
p(x)=(x+3)^2-2.
$$
Then compose to get
$$
p(p(p(p(x))))=\big(\big(\big((x+3)^2+1\big)^2+1\big)^2+1\big)^2-2,
$$
and now it's clear that the minimum argument is $x=-3$, and the minimum value $p(p(p(p(-3))))=23$.
A: Notice 


*

*$p(x) = (x+3)^2 - 2$ is strictly increasing on $[-3,\infty)$ 

*$p([-3,\infty)) \subset p(\mathbb{R}) = [-2,\infty) \subset [ -3, \infty)$. 


This implies for any $a \in [-3,\infty)$, $p([a,\infty)) = [p(a),\infty)$.
As a result, for any $n >  0$, 
$$\begin{align} & p^{\circ n}(\mathbb{R}) \stackrel{def}{=}
\underbrace{p(p(\cdots p}_{n \text{ times}}(\mathbb{R})\cdots) 
= p^{\circ n-1}([-2,\infty)) = [ p^{\circ n-1}(-2), \infty )\\
\implies & \min_{x \in \mathbb{R}} p^{\circ n}(x) = p^{\circ n-1}(-2)
\end{align}
$$
To find the minimum of $n$-fold iteration of $p$, one just need to compute the $(n-1)$-fold iteration of $p$ at $-2$. In particular, 
$$\begin{align}
\min_{x\in\mathbb{R}} p^{\circ 4}(x) &= p^{\circ 3}(-2)\\
&= p^{\circ 2}((-2+3)^2 - 2)
= p^{\circ 2}(-1)\\
&= p( (-1+3)^2-2) = p(2)\\
&= (2+3)^2-2 = 23
\end{align}$$
A: $p^{(4)}=p^{(2)}\circ p^{(2)}$. 
The range of $p^{(2)}$ is $[-1,\infty)$, $p^{(2)}$ is monotonically increasing over $[-1,\infty)$, and attains its minimum at $-1$. This minimum is $p^{(2)}(-1)=23$, which rewrites again as $p^{(4)}(-3)=23$.
A: Let's denote that $min\ p_n(x)=a_n$, where $n$ is number of nests/folds of $p(x)$. For example, $p_1(x)=p(x), p_2(x)=p(p(x))$.
Now it's easy to show that,
$a_n=(a_{n-1}+3)^2-2, a_1=-2, n \geq 2$.
For example, $min\ p_1(x)=a_1=0-2=-2 \\min\ p_2(x)=a_2=(a_1+3)^2-2=-1\\min\ p_3(x)=a_3=(a_2+3)^2-2=2\\min\ p_4(x)=a_4=(a_3+3)^2-2=23\\min\ p_5(x)=a_5=(a_4+3)^2-2=674\\min\ p_6(x)=a_6=(a_5+3)^2-2=458,327$.
It increases too fast.
