Example 3.31 Boyd's Book:
My question is, how can we determine that the function $f$ is quasiconcave/quasiconvex?
Similarly, for $f(x_1, x_2) = x_1/x_2 $ on $R^2_{++}$ is given as both quasiconvex and quasiconcave. However, I cant see why? Can anyone please explain, how to find whether a convex function is quasiconcave or not? Thanks.
Here is one way to prove that the proposed function is quasiconcave. (There are alternative approaches.) Using your cue, we need to check that the upper level sets $\{ \mathbf{x}: f(\mathbf{x} \ge \alpha \}$ are convex for all $\alpha$. Given $f(x_1,x_2) = x_1x_2$, this is equivalent to checking that the upper level sets $$\{ \mathbf{x} \in \mathbf{R}^2_+: x_1x_2 \ge \alpha \} = \{ \mathbf{x} \in \mathbf{R}^2_+: x_2 \ge \alpha / x_1\}$$ are convex. This follows because each set $\{ \mathbf{x} \in \mathbf{R}^2_+: x_2 \ge \alpha x_1\}$ is the intersection of two convex sets (the region above the hyperbole $x_2 = \alpha / x_1$ and the positive orthant) in the $(x_1,x_2)$-plane.
Consider now the function $f(x_1,x_2) = x_1/x_2$. Each upper level sets $\{ \mathbf{x} \in \mathbf{R}^2_+: x_1 \ge \alpha x_2\}$ is the intersection of two convex sets (namely, the half-space above the line $x_2 = \alpha x_1$ and the positive orthant) in the $(x_1,x_2)$-plane and thus is a convex set. Then $f$ is quasiconcave. Likewise, the lower level sets $\{ \mathbf{x} \in \mathbf{R}^2_+: x_1 \le \alpha x_2\}$ are also convex for all $\alpha$ and thus $f$ is also quasi convex.