What is the possible maximum/minimum height for a $29$ nodes AVL tree?
Maximum:
We've learned in class that $h\le \phi \log_2n$. Where $\phi\approx 1.4404$. So if we calculate for $n=29$ we get that $h \le 7-\varepsilon$. Meaning, the maximum height is $6$, but the question asks for the possible maximum. Is that still right?
Minimum:
To get the minimum we need to build a perfect tree. A perfect tree has $2^h-1$ nodes. Hence,
$2^h-1 \ge 29 \implies 2^h \ge 30$. Hence, $h=5$.
Is that right?
Correct answer for the minimum, but wrong reasoning for both.
To show that the minimum height for $n$ nodes is $x$, you have to first show that some valid AVL tree with $n$ nodes has height $x$, and then show that any other AVL tree with $n$ nodes cannot have smaller height. Both parts should be easy.
Similarly, to show that the maximum height for $n$ nodes is $y$, you have to first show that some valid AVL tree with $n$ nodes has height $y$, and then show that it is the best. You have quoted an upper bound, but it does not mean that the nearest integer less than it is achievable. For that you need to explicitly construct a tree to prove it.