# Upper and Lower Bounds

## Key Questions

Best way is to find it from the graph of the function.

#### Explanation:

The range of a function can be found in many ways , best way and a concrete way is by plotting its graph and determining.

How you draw the graph depends on you, a generic way is by differentiating the function to get the critical points or the points of maxima or minima . Then you double differentiate the function to see if the critical point is a maxima or minima by putting the critical point in the double differentiated function. If the output is positive then is a minima , if its negative then its a maxima and if its zero then the test fails,we have to go for high derivative tests. The global maxima points give you the upper bound.

Also we can determine the point of inflection for the given curve to determine the convexity or concavity changes of the given function to make the graph more precise .

Thus, having plotted the graph we can tell the upper bound of the function by merely looking at the maximum value of y for all values of x.

• A function $f$ is bounded in a subset $U$ of its domain if there exist constants $M , m \in \mathbb{R}$ such that

$m \le f \left(x\right) \le M ,$ for all $x \in U .$

For example,

1. $f \left(x\right) = \sin \left(x\right)$ is bounded in $\mathbb{R}$ because

$- 1 \le \sin \left(x\right) \le 1 ,$ for all $x \in \mathbb{R}$.
2. $f \left(x\right) = {x}^{2}$ is bounded in $\left[0 , 1\right]$ because

$0 \le {x}^{2} \le 1 ,$ for all $x \in \left[0 , 1\right] .$

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.

• If real numbers $L$ and $U$ satisfy the inequalities

$L \le f \left(x\right) \le U$,

for all $x$ in the domain of $f$, then $L$ is a lower bound of $f$ and $U$ is an upper bound of $f$.

I hope that this was helpful.