Local maximum and local minimum

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Derivatives can be used to find a function's local extremal values (i.e., local maximum or minimum values).

If \(f(x)\) is a continuous function defined on a closed interval \([a,b]\), then it always has a maximum value and a minimum value (c.f. the Extreme Value Theorem). Finding a maximum or minimum of a function on its entire domain is a global problem.

On the other hand, examining continuity or differentiability around \(x=a\) is a local problem in the sense that it only concerns the neighbor of the point \(x = a\). We may also define the notions of local maxima and local minima.



Definition (Local maximum and local minimum)

Let \(f(x)\) be a function defined on an interval \(I\) and let \(a \in I\).

  1. \(f(a)\) is said to be a local maximum value of the function \(f(x)\) if there exists  \(\delta > 0\) such that, for all \(x\in (a - \delta, a + \delta) \cap I\), \(x \neq a\) implies \(f(x) < f(a)\). \(x= a\) is said to be a local maximum point if \(f(a)\) is a local maximum value.
  2. \(f(a)\) is said to be a local minimum value of the function \(f(x)\) if there exists  \(\delta > 0\) such that, for all \(x\in (a - \delta, a + \delta) \cap I\), \(x \neq a\) implies \(f(x) > f(a)\). \(x= a\) is said to be a local minimum point if \(f(a)\) is a local minimum value.
We use the term extreme value to mean either a local maximum value or a local minimum value.

Remark. If a local maximum value is actually a maximum value, then the maximum value is called a global maximum value. A global minimum value is similarly defined. Note that a global maximum (minimum) value is a local maximum (minimum) value, but not vice versa. □

Example. The constant function \(f(x) = c\) (\(c\) is a constant) has the maximum value \(c\) and the minimum value \(c\), but it does not have a local maximum value or local minimum value. □

Example. Let \(f(x) = (x^2 - 1)(x^2 - 4)\) (draw the graph!). \(f(0) = 4\) is a local maximum value. In fact, take \(\delta = \sqrt{5} > 0\), then for all \(x\) such that \(0 < |x| < \delta\), we have
\[0  < x^2 < \delta^2\]
(i.e., \(x^2\) is strictly positive) and
\[-5  < x^2 - 5 < \delta^2 - 5 = 0\]
(i.e., \(x^2 - 5\) is strictly negative) so that
\[f(x) = x^2(x^2 - 5) + 4 < 4 = f(0).\]
This means that \(f(0) > f(x)\) for all \(x \in (-\delta, +\delta), x \neq 0\). That is, \(f(0)\) is a local maximum value (check the above definition again!). □

Definition (Stationary/critical point)

Let \(f(x)\) be a differentiable function. \(x=a\) is said to be a critical point or stationary point if \(f'(a) = 0\). In this case, \(f(a)\) is called a critical value or stationary value.

Theorem

If a differentiable function \(f(x)\) has a local extremum value (i.e., either a local maximum or a local minimum value) at \(x = a\), then \(x = a\) is a stationary point.
Proof. We prove only for the local maximum case (the local minimum case is similar).
Suppose \(f(x)\) takes a local maximum value at \(x=a\). Then there exists \(\delta > 0\) such that, for all \(x\in (a - \delta, a + \delta)\), \(f(x) \leq f(a)\). Since \(f(x)\) is differentiable at \(x = a\) (by assumption), we have
\[ \begin{eqnarray} f'(a) &=& \lim_{x\to a}\frac{f(x) - f(a)}{x - a}\\ &=&\lim_{x\to a+0}\frac{f(x) - f(a)}{x - a}\\ &=&\lim_{x\to a-0}\frac{f(x) - f(a)}{x - a}. \end{eqnarray} \]
  • If \(0 < a - x < \delta\) (i.e., \(x \in (a - \delta, a)\)), then \(\frac{f(x) - f(a)}{x - a} \geq 0\) so, by taking the limit as \(x \to a-0\), we have \(f'(a)\geq 0\). 
  • If \(0 < x - a < \delta\) (i.e., \(x\in (a, a+\delta)\)), then \(\frac{f(x) - f(a)}{x - a} \leq 0\) so, by taking the limit as \(x \to a+0\), we have \(f'(a) \leq 0\). 
Therefore, we have \(f'(a) = 0\). ■
Remark. The converse of this theorem is not true. That is, \(f'(a) = 0\) does not imply \(f(a)\) is an extremum. For example, \(f(x) = x^3\) has \(f'(0) = 0\), but \(f(0)\) is neither a local maximum value nor a local minimum value (draw the graph!). □

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