A function $f(x)$ is said to have a relative maximum at $x = x_0$ if $f(x_0) \geq f(x)$ for all $x$ in some open interval containing $x_0$,that is, if the value of $f(x_0)$ is greater than or equal to the values of $f(x)$ at all nearby points. The neighborhood is $0 <|x-c|< \delta$.

My question: Consider a function $f(x)=x$. Consider two points, $x=1$ and $x=1.1$, we will get $f(1.1) \geq f(1)$ but again, $f(1.1) \geq f(1.2)$, and so on. So which is the maxima of the function $f(x)=x$? Similar argument for the minima.

Do we have $n$ number of maxima and minima for any function?

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    $\begingroup$ To use $\TeX$, enclose it in dollar signs (single for inline, double for display). You need backslashes in commands like \geq, but not elsewhere, e.g. before f. $\endgroup$ – joriki Feb 13 '12 at 14:56

The conclusion you seem to be making is not correct; just knowing that $f(1.1)>f(1)$, for example, does not tell you that $f(1.1)$ is a local maximum value. To show that $f(a)$ is a local maximum value, you'd have to prove that $f(x)\le f(a)$ for every $x$ in the domain of $f$ sufficiently close to $a$. Here, you cannot do that; $f$ has no local minimum or maximum values by your implied argument.

Note that the definition you give is for relative maximum and minimum values. Whereas you seem to be referring to global maximum and minimum values in the sequel.

A local maximum value $f(x_0)$ is a the largest value the function attains in a neighborhood of $x_0$. Note, the definition said that $f(x_0)$ is greater than or equal to the values of $f(x)$ at all nearby points $x$.

A global maximum value $f(x_0)$ must satisfy $f(x_0)$ is greater than or equal to the values of $f(x)$ for all $x$.

There are similar definitions for local and global minimum values.

These concepts also take into account the domain of the function.

For instance, if you restrict $f(x)=x$ to have domain $[0,1.1]$, say, then $f(0)=0$ will be both a local and a global minimum value and $f(1.1)=1.1$ will be both a local and a global maximum value. To be precise, we would say $f$ has both a local and global maximum value of $f(1.1)=1.1$ over $[0,1.1]$; and, $f$ has both a local and a global minimum value of $f(0)=0$ over $[0,1.1]$.

If you consider $f(x)=x$ to have domain $\Bbb R$, as I assumed in the first paragraph, then it would have no local or global maximum values by your observations. It would also have no local or global minimum values.

  • $\begingroup$ Thanx for the reply, but my confusion persists, for a real line, any function will have n number of points on it so we will have n number of intervals enclosing each of the n points and sufficiently close to each of the n points, and for each of these n intervals we will have one value greater than another value, so do we arrive at the conclusion that we have n maxima and minima? $\endgroup$ – Vikram Feb 13 '12 at 15:49
  • $\begingroup$ @Vikram No. You have to $fix$ the interval you are working over at the outset. Say we consider $[0,10]$. Now take a point $a$ in this interval. $f(a)$ will be a local maximum value if and only if there is a (small) interval around $a$ such that the values of $f$ for $every$ point in that interval are at most $f(a)$. Note that this includes points on "both sides" of $a$. If you look at the graph of the function, it would have a "hump" at $(a,f(a))$. $\endgroup$ – David Mitra Feb 13 '12 at 15:58
  • $\begingroup$ @Vikram It is true that $f(x)=x$ has the maximum value $1$ over $[0,1]$ and has the maximum value 2 over $[0,2]$, etc... But notice I said "maximum value over $[0,1]$, maximum value over $[0,2]$, etc...". These different maximum values are referring to $different$ intervals. $f(1)=1$ is $not$ a maximum value over $[0,2]$ (look at the graph), but is is a maximum value over the interval $[0,1]$. $\endgroup$ – David Mitra Feb 13 '12 at 16:04
  • $\begingroup$ Thanx David, I think you clarified my doubt, now I will solve some problems to check my understanding, but I feel your explanation does fill the gaps in my understanding about this topic. $\endgroup$ – Vikram Feb 13 '12 at 16:36

The function $f(x) = x$ has no absolute or relative maxima/minima. No matter what value you give me for the relative maxima (respectively, minima), I can find a larger value (respectively, smaller value) nearby obtained by the function.

To answer the second part, yes you can find a function which has any number of prescribed maxima/minima. You can draw a sawtooth function with as many "high points" or "low points" as you wish, for example.


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