# General method to find inf, sup, maxs and mins of a function

Could someone explain how to find inf, sup, max and min values of a function (real-valued functions of real variable, generally continuous/differentiable, with some possible points of discontinuity)?
Some examples: $$y = \frac{x}{2}\sqrt{\frac{\log{x} + 1}{\log{x} - 1}}$$ $$y = \frac{x^2}{1+\log|x|}$$ $$y = e^{x-|x^2-x|}$$ $$y = e^{\arctan{\frac{1}{x-1}}}$$ $$y = \left(\frac{1}{2}\right)^{\frac{|x|}{1+x}}$$ $$y = x\log{\frac{1}{|x|}}$$

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What kind of function? What context? Calculus? – Arturo Magidin Jul 3 '12 at 18:23
In what context are you studying functions? What kind of functions are these? "Function study" is like answering "What book are you reading" by saying "It's a book that I'm reading." No information has been provided. Without context, the question is too broad to be answered. Different types of situations call for different kinds of tools. – Arturo Magidin Jul 3 '12 at 18:27
Can you give a few examples of functions for which you want to find these things? – roninpro Jul 3 '12 at 18:35
– user2468 Jul 3 '12 at 18:35
So you are talking about "mathematical analysis". "What kind of functions" refers to what kinds of inputs and outputs your functions take, what their properties are (continuous? differentiable? etc) Do you have examples of the kinds of questions you are talking about? (And if your question comes from doing a course at your University, why not say what it is about the course that is confusing you? – Arturo Magidin Jul 3 '12 at 18:35

If you are dealing with real-valued functions of real-variable, then generally you want to:

1. Determine the domain of the function.

2. Determine where the function is continuous.

For intervals of continuity, any local extreme will occur at either an endpoint or a critical point (points where the derivative either does not exist, or is equal to $0$). Local extremes may also occur at points of discontinuity. You can also use the derivative to determine intervals on which the function is increasing or decreasing, which will of course help you locate extreme points.

So generally, you want to determine those intervals of continuity, and work each of them using the derivative and the sundry tests that exist for checking critical points to determine if they are local extremes or not.

Example. Let's take your last function: $$y = x\log{\frac{1}{|x|}}.$$ First, the domain of this function is all $x\neq 0$. The function is continuous at every point of its domain, so we will want to look at what happens on $(0,\infty)$ and on $(-\infty,0)$. In addition, we want to understand what happens as the function approaches the "endpoints" of those intervals ($0$ on either side, $\infty$, and $-\infty$); this is done using limits.

First, as $x\to \infty$, we have the function $$y = x\log\frac{1}{x} = -x\log x.$$ As $x\to\infty$, $\log x\to\infty$; multiplied by a function that goes to $-\infty$, the product will go to $-\infty$. That is, $$\lim_{x\to\infty}x\log\frac{1}{|x|} = -\infty.$$ Moreover, if we look at the derivative, $$y' = -\log x - x\left(\frac{1}{x}\right) = -\log x- 1.$$ This will be positive if $\log x \lt -1$, if and only if $x\lt \frac{1}{e}$; and negative if $x\gt \frac{1}{e}$. We have a critical point at $x=\frac{1}{e}$.

Since $y$ is increasing on $(0,\frac{1}{e})$ and decreasing on $(\frac{1}{e},\infty)$, the function has a local maximum at $\frac{1}{e}$ and no other local extremes on $(0,\infty)$.

We can also verify that $$\lim_{x\to 0^+}-x\log x = 0,$$ e.g., using L'Hopital's Rule: \begin{align*} \lim_{x\to 0^+}-x\log x &= \lim_{x\to 0^+}\frac{-\log x}{\frac{1}{x}}\\ &= \lim_{x\to 0^+}\frac{-(1/x)}{-(1/x^2)}\\ &= \lim_{x\to 0^+} x\\ &= 0. \end{align*} Thus, there are no minimum or local minimum values.

We can do the same analysis on $(-\infty,0)$, or we can note that the function is odd (exploit symmetry whenever possible): $$y(-x) = (-x)\log\frac{1}{|-x|} = -x\log\frac{1}{|x|} = -y(x).$$ So we can just reflect about the origin to see what happens on $(-\infty,0)$. The function approaches $0$ from below as $x\to 0^-$; the function is increasing on $(-\frac{1}{e},0)$ and decreasing on $(-\infty,-\frac{1}{e})$; there is a local minimum at $x=-\frac{1}{e}$, and $$\lim_{x\to-\infty}x\log\frac{1}{|x|} = \infty.$$

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Thank you very much! – Overflowh Jul 4 '12 at 15:28

There is no general method that works for an arbitrary function. There are functions for which the sup/inf/etc... cannot be obtained analytically, but only its existence can be proved. For other functions, it is unknown whether such a bound even exists.

However, I assume you are talking about simple differentiable functions of the genre that you gave in the question. For these kind of functions, there are a few steps that one usually preforms.

1. Find the singularities - $x$'s for which the function goes to $\pm\infty$. If you found one, your function is unbounded from above/below/both.

2. Find critical points - $x$'s for which $\frac{df}{dx}=0$. These are (usually) local maxima / minima. Many times one can obtain analytically all these points and then the behavior of the function between these points can be deduced.

3. Examine the behavior of your function in the boundaries of the domain.

If you succeeded in doing all 3 steps analytically, you inevitably have the maxima/minima of your function.

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Thanks a lot for the step by step way :) – Overflowh Jul 4 '12 at 15:28