# How do I find the lower bound of a function?

Jun 23, 2018

See below.

#### Explanation:

First of all, let's get rid of infinities: a function can tend to $\setminus \pm \setminus \infty$ either at the extreme points of its domain or because of some vertical asymptote.

So, you should first of all check

${\lim}_{x \setminus \to {x}_{0}} f \left(x\right)$

for every point ${x}_{0}$ at the boundary of the domain. For example, if the domain is $\left(- \setminus \infty , \setminus \infty\right)$, you should check

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right)$

If the domain is like $\setminus m a t h \boldsymbol{R} \setminus \setminus \setminus \left\{2 \setminus\right\}$ you should check

${\lim}_{x \setminus \to \setminus \pm \setminus \infty} f \left(x\right) , \setminus q \quad {\lim}_{x \setminus \to {2}^{\setminus} \pm} f \left(x\right)$

and so on. If any of these limits is $- \setminus \infty$, the function has no finite lower bound.

Else, you can check the derivative: when you set $f ' \left(x\right) = 0$, you will find points of maximum of minimum. For every $x$ which solves $f ' \left(x\right) = 0$, you should compute $f ' ' \left(x\right)$. If $f ' ' \left(x\right) > 0$, the point is indeed a minumum.

Now, in the most general case, you have a collection of points ${x}_{1} , \ldots , {x}_{n}$ such that

$f ' \left({x}_{i}\right) = 0 , \setminus q \quad f ' ' \left({x}_{i}\right) > 0$ for every $i = 1 , . . , n$

Which means that they are all local minima of your function. The lower bound of the function, i.e. the global minimum, will be the smallest image of those points: you just need to compare

$f \left({x}_{1}\right) , \ldots , f \left({x}_{n}\right)$, and choose the smallest one.