This is from a test (I'm not a high school student) given to rising high school juniors. The problem was designed to take less than 20 minutes, preferably 5-15. Judging from its source, this is not a trick question, and I am pretty sure all information given must be used.
I already know its answer, but the difficulty is that we cannot use continuity, sequence, differentiation or integration.
Let $f(x)$ be a quadratic polynomial with a positive leading coefficient. Let $g(x)$ be $ 1 - \frac{2}{x-5}$ with the domain $x < 5$. For any real number $t < 3$, let $h(t)$ be the minimum of $f(g(x))$ for $t \leq x \leq t+2$. Suppose also that $h(t)= f(g(t+2))$ when $t < 1$ and $h(t) = 6$ for $1 \leq t < 3$ and that $h(-1) =7$. Find $f(5)$.
It suffices to find an explicit formula for $f$, which is the direction I am heading at.
The slight difficulty here is that we cannot use continuity of $h$ (easily derived from the continuity of $f(g(x))$ and the fact that $[t,t+2]$ is compact; please correct me if I'm wrong). If we are allowed to use continuity, then $6= h(1) = f(g(3))=f(2)$ is the vertex of $f$.
We know from $h(-1) =7$ that $f(\frac32) = 7$. So we know the positions of the vertex and another point on the graph of the quadratic $f$ as well as the fact that $f$ has a positive leading coefficient, so we can get an explicit formula of $f$ (I won't state it here since the solution is purely computational).
Again, I'm looking for a solution that does not use continuity.
Any help would be greatly appreciated.
As $g$ is increasing on whole domain (and it's an infinite interval), composition $f(g(x))$ increases where $f(x)$ increases. There is the "tip", say $g(x_0)$ s.t. for all smaller numbers, $f$ is decreasing and for all bigger, $f$ is increasing (basic properties of quadratics, and uses the fact that first coefficient of $f$ is positive). That means that if $x_0>t+2$, we know that $f(g(x))$ is strictly decreasing on $[t, t+2]$, therefore minimum is at right end of the interval. If $t \leq x_0 \leq t+2$, then minimum of $f(g(x))$ is at $x_0$. This in particular means that $x_0=3$, so the lowest point of $f$ is at $g(3)$.
So you have $7=h(-1)=f(g(1))=f(3/2)$ and $f(g(3))=6$, $-b/a=g(3)$, where $f(x)=ax^2+bx+c$.