SAT 1000 problems - P2

If one graphs $f(x)$ out, one finds that:

$f(-1) = f(4) = 0$ and $f(0) = f(3) = 1$

and that $lim_{x \rightarrow -\infty} f(x) \rightarrow 1$ is a horizontal asymptote. If there exists real $p \neq q \neq r$ such that $f(p) = f(q) = f(r)$ , then two necessary and sufficient conditions are:

$0 < f(p), f(q), f(r) < 1$ (i)

$p \in (-\infty, -1); q \in (-1, 0); r \in (3, 4)$ (ii)

As $f(x) \rightarrow 0$ , this corresponds to $p, q \rightarrow -1, r \rightarrow 4$ . This computes to an upper bound of $2^{p} + 2^{q} + 2^{r} = 2^{-1} + 2^{-1} + 2^4 = 17.$

As $f(x) \rightarrow 1$ , this corresponds to $p \rightarrow -\infty, q \rightarrow 0, r \rightarrow 3$ . This computes to a lower bound of $2^{p} + 2^{q} + 2^{r} = 2^{-\infty} + 2^{0} + 2^3 = 9.$

Hence, the desired range of $2^{p} + 2^{q} + 2^{r}$ comes to $\boxed{(9, 17)}.$