Finding $\displaystyle \sup_{x \in [1,\infty)} x e^{-tx}$ when $t \ge 0$

Question

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1. What the problem is asking

For a fixed non‑negative parameter (t) we consider the function

[ f_t(x)=x\,e^{-t x}, \qquad x\in[1,\infty). ]

We have to determine the supremum (the least upper bound) of the set

[ {f_t(x):x\in[1,\infty)} ]

for each possible value of (t\ge 0).
In other words: “How large can the product (x e^{-t x}) become when we are allowed to choose any (x\ge 1)?”

2. Detailed solution

2.1 The case (t=0)

If (t=0) then

[ f_0(x)=x e^{0}=x . ]

On the interval ([1,\infty)) the function is just the identity, so it grows without bound:

[ \sup_{x\ge 1} f_0(x)=\sup_{x\ge 1} x =+\infty . ]

2.2 The case (t>0)

2.2.1 First derivative

[ f_t(x)=x e^{-t x}, \qquad t>0. ]

Differentiate (product rule):

[ f’_t(x)=e^{-t x}+x(-t)e^{-t x}=e^{-t x}(1-tx). ]

Because the exponential factor is never zero, the critical points are the solutions of

[ 1-tx=0\quad\Longrightarrow\quad x=\frac{1}{t}. ]

2.2.2 Where does the critical point lie?

Our domain is ([1,\infty)).
The point (x=\frac{1}{t}) belongs to this interval iff (\frac{1}{t}\ge 1), i.e. iff

[ 0<t\le 1 . ]

Thus we have two sub‑cases:

(a) (0<t\le 1) – the critical point is inside the domain.
(b) (t\ge 1) – the critical point is to the left of the domain; on ([1,\infty)) the derivative never changes sign.

2.2.3 Behaviour for large (x)

Regardless of (t>0),

[ \lim_{x\to\infty} x e^{-t x}=0 . ]

A quick proof with L’Hôpital:

[ \lim_{x\to\infty}\frac{x}{e^{t x}} =\lim_{x\to\infty}\frac{1}{t e^{t x}}=0 . ]

Hence the function cannot have its supremum at (+\infty).

2.2.4 Sub‑case (a): (0<t\le 1)

Here (x_0=1/t) belongs to the interval, and the derivative changes sign:

For (1\le x<1/t), we have (1-tx>0) → (f’_t(x)>0) (increasing).
For (x>1/t), we have (1-tx<0) → (f’_t(x)<0) (decreasing).

Thus (x_0) gives the global maximum on ([1,\infty)).
Its value is

[ f_t!\left(\frac{1}{t}\right)=\frac{1}{t}\,e^{-t\cdot\frac{1}{t}}= \frac{1}{t}\,e^{-1}= \frac{1}{t e}. ]

Because the limit at infinity is (0) and (f_t(1)=e^{-t}<\frac{1}{t e}) for (0<t\le1), the supremum equals this value:

[ \boxed{\displaystyle \sup_{x\ge 1} f_t(x)=\frac{1}{t e}\qquad (0<t\le 1)}. ]

2.2.5 Sub‑case (b): (t\ge 1)

Now (1/t\le 1), so the critical point is outside the domain.
For every (x\ge 1),

[ 1-tx\le 1-t\le 0\quad\Longrightarrow\quad f’_t(x)\le 0 . ]

Hence (f_t) is strictly decreasing on ([1,\infty)).
The largest value is therefore attained at the left endpoint (x=1):

[ f_t(1)=e^{-t}. ]

Since the limit at infinity is (0), we obtain

[ \boxed{\displaystyle \sup_{x\ge 1} f_t(x)=e^{-t}\qquad (t\ge 1)}. ]

2.2.6 Summary for (t>0)

Putting the two sub‑cases together,

[ \sup_{x\ge 1} x e^{-t x}= \begin{cases} \dfrac{1}{t e}, & 0<t\le 1,\[6pt] e^{-t}, & t\ge 1 . \end{cases} ]

Notice that the two formulas coincide at (t=1) because (\frac{1}{1e}=e^{-1}).

2.3 Final answer (all (t\ge 0))

[ \boxed{ \sup_{x\in[1,\infty)} x e^{-t x}= \begin{cases} +\infty, & t=0,\[6pt] \dfrac{1}{t e}, & 0<t\le 1,\[8pt] e^{-t}, & t\ge 1 . \end{cases}} ]

The supremum is actually a maximum in every case (it is attained at (x=\infty) when (t=0), at (x=1/t) when (0<t\le1), and at (x=1) when (t\ge1)).

3. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Ignoring the restriction (x\ge 1) when locating the critical point.	The derivative gives (x=1/t), but if (t>1) this point lies left of the interval, so it cannot be a candidate for the supremum.	After solving (f’(x)=0), always check whether the solution belongs to the given domain.
Assuming the supremum is always at the critical point.	When the critical point is outside the domain the function is monotone on the whole interval, and the extremum occurs at an endpoint.	Examine the sign of (f’(x)) on the interval; if it never changes sign, the extremum is at an endpoint.
Treating (\lim_{x\to\infty} x/e^{t x}) as “(\infty/\infty)” and stopping.	The indeterminate form must be resolved (e.g., by L’Hôpital’s rule) to see that the limit is actually (0).	Apply L’Hôpital (or compare growth rates: exponential dominates any polynomial) to compute the limit.
Missing the case (t=0).	For (t=0) the function reduces to (x), which is unbounded, so the supremum is (+\infty).	Separate the analysis into (t=0) and (t>0) right at the start.
Writing (\sup\lim) instead of (\sup).	“(\sup\lim)” is not a standard notation; we first take the supremum of the function values, not of a limit.	Use the correct notation: (\displaystyle \sup_{x\in[1,\infty)} f_t(x)).

Keeping these points in mind will help you solve similar optimisation‑over‑unbounded‑interval problems without error.

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