A sequence of numbers $A=(a_1, a_2, \dots, a_n)$ is called strictly convex, if there is a $k$, with $1 \leq k \leq n$ so that for all $1 \leq i \leq k-1$ we have $a_i>a_{i+1}$ and for all $k \leq i \leq n-1$ we have $a_i<a_{i+1}$.
Which is the best algorithm that finds the minimum element of the sequence $A$ in time $O(\log n)$?
Use binary search algorithm to find such a $j$ for which $a_{j-1}>a_{j}$ and $a_{j+1}>a_{j}$. Then $j=k$ and $a_j$ is minimum element of the sequence.
If $a_{i-1}>a_{i}>a_{i+1}$ you know that $i<k$.
If $a_{i-1}<a_{i}<a_{i+1}$ you know that $i>k$.
You can adapt binary search algorithm to solve this problem.