If f(x) = axe^{-x} is an “M-function” on the interval $(0,+\infty)$

Key takeaway: This is an inequality involving a derivative. The key is to compute $f'(x)$ and test whether the condition can hold for every positive $x$.

Given

$f(x)=axe^{-x},$

first compute the derivative:

$f'(x)=a e^{-x}(1-x).$

The condition for an M-function is

$f'(x)>f(x) \quad \text{for all } x>0.$

Substitute the formulas:

$a e^{-x}(1-x) > a x e^{-x}.$

Since $e^{-x}>0$ for all $x$, divide both sides by $e^{-x}$:

$a(1-x)>ax.$

Bring terms together:

$a(1-2x)>0 \quad \text{for all } x>0.$

But $1-2x$ changes sign on $(0,+\infty)$:

• if $0<x<\tfrac12$, then $1-2x>0$
• if $x>\tfrac12$, then $1-2x<0$

So no fixed real number $a$ can make $a(1-2x)>0$ true for every $x>0$.

Therefore, the set of real values of $a$ is

$\boxed{\varnothing}$

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Pitfalls the pros know 👇 A frequent mistake is to check the inequality at only one point. The condition says it must hold for all $x$ in $(0,+\infty)$. Another mistake is dividing by $a$ without knowing its sign; here the sign issue is exactly why no real $a$ works.

What if the problem changes? If the interval were restricted to $\left(0,\tfrac12\right)$, then $1-2x>0$ throughout that interval, so any $a>0$ would work. If the interval were $(\tfrac12,+\infty)$, then any $a<0$ would work.

Tags: derivative inequality, function classification, parameter range