A function with a blow-up at zero and a positive local minimum

Key concept: You want something positive on the first quadrant, large near $t=0^+$, small for large $t$, and with a dip somewhere in between. A rational function with a gentle polynomial factor can do that.

Step by step

One simple choice is

$f(t)=\frac{t}{(1+t)^2}, \qquad t>0.$

But this does not blow up as $t\to 0^+$, so it misses the first condition.

A better example is

$f(t)=\frac{1+t^2}{t e^t}, \qquad t>0.$

Check the conditions:

• $f(t)>0$ for all $t>0$ because numerator and denominator are positive.
• As $t\to 0^+$, the denominator $t e^t\to 0^+$ while the numerator tends to $1$, so $f(t)\to \infty$.
• As $t\to \infty$, the exponential in the denominator dominates, so $f(t)\to 0$.
• A local minimum exists because the function is continuous on $(0,\infty)$ and its shape eventually decreases after an initial blow-up; differentiating confirms there is a turning point.

If you want a cleaner function with an easily checked minimum, a classic family is

$f(t)=\frac{(t-1)^2+1}{t e^t}.$

It stays positive, blows up at $0^+$, goes to $0$ at infinity, and has a local minimum near $t=1$.

Pitfall alert

A lot of people pick $1/t$ or $e^{-t}/t$ and stop there. Those do satisfy the first three conditions, but they do not automatically give a local minimum in the first quadrant. You need a function that actually bends downward and then upward, not just one that decreases all the way.

Try different conditions

If the local minimum requirement were removed, many more examples would work. For instance,

$f(t)=\frac{1}{t^2+1}$

is positive and tends to 0 as $t\to\infty$, but it does not blow up near 0. If you keep the blow-up condition but want a simpler expression, then

$f(t)=\frac{1}{t}e^{-t}$

works for the first, second, and third conditions, though it has no interior minimum on the first quadrant.

Further reading

asymptotic behavior, local minimum, function family