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I was trying to study the function $f(x)=1-xe^{-x}$, and for studying the sign I came to the expression $1-xe^{-x} \geq 0$ which can be expressed as $\frac{1}{x} \geq e^{-x}$ or as $e^{x} \geq x$ . But these two inequalities return two different solutions because the first is true only when $x\gt0$ while the second is true $\forall x \in \mathbb{R}$, which is the correct solution. Why this happen?

Why is the expression: $\frac{1}{x} \geq e^{-x} $ is different then the expression: $e^{x} \geq x$ ?

Noa Even
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Edoardo Grassi
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4 Answers4

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You need to be careful when you multiply inequalities by $x$ as multiplying with negative numbers reverses the sign. That's why the first one is only true for positive numbers while you "made" the second one always true by multiplying with $x$ (which is negative if $x < 0$). The same happens with $x \geq 0$ and $x^2 \geq 0$.

EDIT: In fact, you did this twice. First you devided by $x$ and then you multiplied by $x$. Hence the function is always positive (the two mistakes cancel).

Klaus
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"I came to the expression $1-xe^{-x} \geq 0$ which can be expressed as $\frac{1}{x} \geq e^{-x}$ or as $e^{x} \geq x$"

They only express the same thing if $x>0$.

drhab
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Hint:

Notice that $f: 1 - xe^{-x}$ is defined for $f: \mathbb{R} \to \mathbb{R}$.

also, $\frac{1}{x} \geq e^{-x}$ is defined for every $x \in \mathbb{R} \setminus \{0\} $

but $e^ x \geq x$ is defined for every $x \in \mathbb{R}$

Noa Even
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  • So for avoiding this mistake, when reducing an inequality, should I keep the same domain? – Edoardo Grassi Jan 28 '19 at 12:27
  • you should make sure that there exists no $x$ such that expression (1) is defined for but expression (2) isn't defined for.

    for this one, as instance, for $x =0$ expression (1): $\frac{1}{x} \geq e^{-x}$ isn't define for,

    but for $x =0$, expression (2): $e^x \geq x $ IS DEFINED FOR

     – Noa Even Jan 28 '19 at 13:38
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Rearranging $1-xe^{-x}\ge 0$ gives the equivalent $xe^{-x}\le 1$. For $g\ge 0$, this implies $xe^{-x}g\le g$. The choice $g=e^x$ is non-negative as required, so we can deduce $x\le e^x$. But if you try to instead use $g=1/x$ to obtain $e^{-x}\le 1/x$, the problem is this $g$ isn't non-negative for all $x$.

J.G.
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