What is the exact difference between $\arg\max$ and $\max$ of a function?

Is it right to say the following?

$\arg\max f(x)$ is nothing but the value of $x$ for which the value of the function is maximal. And $\max f(x)$ means value of $f(x)$ for which it is maximum.

More precisely,

$\arg \max$ returns a value from the domain of the function and $\max$ returns from the range of the function?

  • 1
    $\begingroup$ $arg max - max=arg$ so the difference is the argument. $\endgroup$ Feb 23, 2013 at 14:56

4 Answers 4


Yes, you are right. $\max f(x)$ is the maximum value (if it exists) of $f(x)$ as $x$ varies through some domain, while $\arg\max f(x)$ is the value of $x$ at which this maximum is attained.

However, there could be more than one $x$ value which gives rise to the maximum $f(x)$, in which case $\arg\max f(x)$ would be this set of values of $x$ instead.


Suppose $f(x) = 100 - (x-6)^2$.

Then $\max\limits_x f(x) = 100$ and $\operatorname*{argmax}\limits_x f(x) = 6$.

  • 2
    $\begingroup$ such simple example but explains really well $\endgroup$
    – user634512
    Feb 2, 2020 at 14:34

$\arg \max_x f(x) = f^{-1} \{ \max_x f(x) \}$.

For example, $\max_x \cos x = 1$, $\arg \max_x \cos x = \{ n 2 \pi\}_{n \in \mathbb{Z}}$.


Yes. $\arg\max f(x)$ is the function argument $x$ at which the maximum of $f$ occurs, and $\max f(x)$ is the maximum value of $f$. Hence $\max f(x)=f\left(\arg\max f(x)\right)$.


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