Question

Let $f, g : \mathbb{R} \to \mathbb{R}$ be defined by \[ f(x) = x e^{-x} \quad \text{and} \quad g(x) = x|x|. \] Then, on $\mathbb{R}$,

• A. both $f$ and $g$ are convex
• B. $f$ is convex and $g$ is not convex
• C. $f$ is not quasiconvex and $g$ is quasiconvex
• D. neither $f$ nor $g$ is quasiconvex

Hint:

A function can fail to be convex but still be quasiconvex if all its sublevel sets are convex — always check the definition, not just curvature.

Answer 1

The Correct Option is B

Step 1: Analyze $f(x) = x e^{-x}$.
First derivative: \[ f'(x) = e^{-x}(1 - x). \] Second derivative: \[ f''(x) = e^{-x}(x - 2). \] For $x<2$, $f''(x)<0$ (concave), and for $x>2$, $f''(x)>0$ (convex). Hence, $f$ is neither globally convex nor concave, and its level sets are not convex — so it is not quasiconvex.
Step 2: Analyze $g(x) = x|x|$.
For $x \ge 0$, $g(x) = x^2$; for $x<0$, $g(x) = -x^2$. Thus, - For $x \ge 0$, $g$ is convex (upward curve). - For $x<0$, $g$ is concave (downward curve). Although not convex overall, $g(x)$ is **quasiconvex**, since the sublevel sets $\{x : g(x) \le c\}$ are convex intervals for all $c$.
Step 3: Conclusion.
$f$ is not quasiconvex, while $g$ is quasiconvex. Hence, option (C) is correct.