Maximum, Minimum, and Inflection Points

Global maximum and minimum points

1.The maximum and minimum of a function $f(x)$ are, respectively, the highest and lowest values the function attains within its domain. Given a function $y = f(x)$ with domain $D,$ a point $x_0 \in D$ is a global maximum point if $f(x_0) \geq f(x)$ for every $x \in D.$ The value $f(x_0) = M$ is the global maximum of the function.

2.Given a function $y = f(x)$ with domain $D,$ a point $x_0 \in D$ is a global minimum point if $f(x_0) \leq f(x)$ for every $x \in D.$ The value $f(x_0) = m$ is the global minimum of the function.

3.If the global maximum and global minimum values of a function exist, they are unique, although each of them may be attained at more than one point of the domain. By Weierstrass's Theorem, if a function is continuous on a closed and bounded interval $[a, b],$ then it attains both a global maximum and a global minimum on that interval.

4.Outside the hypotheses of the theorem, a function may fail to attain a maximum or a minimum, as happens for instance on open intervals or for unbounded functions. In these cases the set of values of the function still has a least upper bound and a greatest lower bound, possibly infinite, which are studied in the entry on supremum and infimum.

Local maximum and minimum points

5.A function can have several local maxima and local minima, points where it attains its highest or lowest value only in a neighborhood, without necessarily being an extreme value over the entire domain.

6.Given a function $y = f(x)$ defined on an interval $[a, b],$ the point $x_0 \in [a, b]$ is a local maximum point if there exists a neighborhood $I$ of the point $x_0$ such that $f(x_0) \geq f(x)$ for every $x$ in the interval $I.$

7.A sufficient criterion for recognizing a local maximum point is based on the sign of the first derivative. Consider a function $y = f(x)$ that is defined and continuous in a neighborhood of the point $x_0,$ and differentiable in the same neighborhood for every $x \neq x_0.$ Suppose that for every $x$ in the neighborhood the derivative satisfies:

$$ \begin{align} f'(x) &> 0 \quad \text{for} \quad x < x_0 \\[6pt] f'(x) &< 0 \quad \text{for} \quad x > x_0 \end{align} $$

8.Under these conditions, $x_0$ is a point of local maximum for the function $f(x).$ The behavior is summarized by the following sign chart:

10.Given a function $y = f(x)$ defined on an interval $[a, b],$ the point $x_0 \in [a, b]$ is a local minimum point if there exists a neighborhood $I$ of the point $x_0$ such that $f(x_0) \leq f(x)$ for every $x$ in the interval $I.$

11.Under the same hypotheses of continuity and differentiability, suppose that for every $x$ in the neighborhood the derivative satisfies:

$$ \begin{align} f'(x) &< 0 \quad \text{for} \quad x < x_0 \\[6pt] f'(x) &> 0 \quad \text{for} \quad x > x_0 \end{align} $$

12.In this case, $x_0$ is a point of local minimum for the function $f(x).$ The corresponding sign chart is:

14.By Fermat's theorem, the local maximum and minimum points of a differentiable function, located within the interior of the domain, are stationary points. This implies that the tangent line at a point of a local maximum or minimum is parallel to the x-axis. In this case, the derivative of the function satisfies $f'(x_0) = 0.$

Proof of the first-derivative criterion

15.The sign criterion stated above follows from Lagrange's Theorem, applied separately on each side of $x_0.$ Consider the case of a local maximum, with $f$ continuous in a neighborhood of $x_0$ and differentiable there for every $x \neq x_0,$ and suppose $f'(x) > 0$ for $x < x_0$ and $f'(x) < 0$ for $x > x_0.$

16.Take a point $x < x_0$ in the neighborhood. On the interval $[x, x_0]$ the function is continuous, and it is differentiable in the interior, so Lagrange's Theorem gives a point $\xi \in (x, x_0)$ such that:

$$f(x_0) - f(x) = f'(\xi)(x_0 - x)$$

17.The point $\xi$ lies to the left of $x_0,$ so $f'(\xi) > 0,$ and the factor $x_0 - x$ is positive. The right-hand side is positive, which gives $f(x) < f(x_0).$

18.Take now a point $x > x_0$ in the neighborhood. On the interval $[x_0, x]$ the same theorem gives a point $\xi \in (x_0, x)$ such that:

$$f(x) - f(x_0) = f'(\xi)(x - x_0)$$

19.This time $\xi$ lies to the right of $x_0,$ so $f'(\xi) < 0,$ while $x - x_0$ is positive. The right-hand side is negative, and again $f(x) < f(x_0).$

20.For every $x \neq x_0$ in the neighborhood we have found $f(x) < f(x_0),$ which is the condition for $x_0$ to be a local maximum. The criterion for a local minimum is obtained by applying the same argument to $-f,$ which reverses the inequalities on the sign of the derivative.

21.

Continuity of $f$ at $x_0$ is what allows Lagrange's Theorem to be applied on intervals having $x_0$ as an endpoint, without assuming that the derivative exists at $x_0$ itself. For this reason the criterion recognizes corners and cusps as extrema, and not only stationary points where $f'(x_0) = 0.$

Upward and downward concavity

22.Let $f(x)$ be a function differentiable at $x_0,$ so that the tangent line to its graph at $x_0$ is defined. We say that the function $f(x)$ is concave upward at $x_0$ if there exists a neighborhood $I$ of $x_0$ such that, for every $x \in I$ with $x \neq x_0,$ the function $f(x)$ takes values greater than those of the line $y = t(x),$ which is the tangent line to the graph of $f(x)$ at $x_0$:

$$f(x) > t(x) \quad \forall x \in I - \{\ x_0 \ \}$$

23.Similarly, we say that the function $f(x)$ is concave downward at $x_0$ if there exists a neighborhood $I$ of $x_0$ such that, for every $x \in I$ with $x \neq x_0,$ the function $f(x)$ takes values less than those of the line $y = t(x)$:

$$f(x) < t(x) \quad \forall x \in I - \{\ x_0 \ \}$$

24.The analytical formulation of these concepts is discussed in the entry on convexity and concavity of functions.

Inflection points and change in concavity

25.An inflection point is a point where the concavity of a function changes.

26.Consider a function $y = f(x)$ defined on an interval $(a, b),$ and let $x_0 \in (a, b)$ be either a point where $f(x)$ is differentiable, or a point where:

$$\lim_{x \to x_0} f'(x) = +\infty \quad \text{or} \quad \lim_{x \to x_0} f'(x) = -\infty$$

27.The point $x_0$ is an inflection point if the function changes concavity at $x_0.$

28.An inflection point is called horizontal if the tangent at the inflection point is parallel to the x-axis. When the tangent is parallel to the y-axis, the inflection point is called vertical. In all other cases it is called oblique.

29.The point $x_0$ is a horizontal inflection point for a function $f(x)$ if $f'(x_0) = 0$ and there exists a neighborhood of $x_0$ in which the sign of $f'(x)$ is the same for every $x \neq x_0$:

31.

The signs in the neighborhood of $x_0$ can be both positive (as in the scheme above) or both negative.

How to calculate the points of local maximum and minimum

32.Given a continuous function, to find the local maximum and minimum points, we analyze the sign of the first derivative. The procedure is:

34.The criterion based on the sign change of $f'$ applies to interior points of the domain. If the function is defined on a closed interval, the endpoints must be examined separately, since only one-sided behavior of the derivative is available there.

35.When the function is twice differentiable, the sign of the second derivative classifies a stationary point, without studying the sign of $f'$ on each side of it. If $f'(x_0) = 0$ and $f''(x_0) \neq 0,$ then $x_0$ is a local maximum when $f''(x_0) < 0$ and a local minimum when $f''(x_0) > 0,$ according to whether the graph is concave downward or concave upward at $x_0.$ This criterion is known as the second derivative test.

36.When $f''(x_0) = 0$ the test is inconclusive, and the nature of $x_0$ follows from the sign of $f'$ or from the higher-order derivatives treated in the entry on higher-order derivatives.

Example 1

37.Let us calculate the local maximum and minimum points of the following function:

$$y = f(x) = x^3 - \frac{1}{2}x^2$$

38.Being a polynomial function, it is continuous and differentiable for all $x \in \mathbb{R}.$ The first derivative of the function is:

$$f'(x) = 3x^2 - x$$

39.We study the sign of the derivative by imposing:

$$3x^2 - x > 0$$

40.Passing to the associated equation and factoring out $x,$ we obtain:

$$3x^2 - x = 0 \implies x(3x -1) = 0$$

41.The equation is satisfied for $x = 0$ and $x = \dfrac{1}{3}.$ Returning to the inequality, since the parabola associated with $3x^2 - x$ opens upward, we obtain that $f'(x) > 0$ for $x < 0$ and for $x > \dfrac{1}{3}.$

42.The sign chart shows that the function is increasing for $x < 0,$ decreasing for $0 < x < \dfrac{1}{3},$ and increasing again for $x > \dfrac{1}{3}$:

44.For $x = 0$ the function takes the value $f(0) = 0^3 - \dfrac{1}{2} \cdot 0^2 = 0.$ The point $(0,0)$ is therefore a local maximum. For $x = \dfrac{1}{3},$ the function takes the value:

$$ \begin{align} f\left( \dfrac{1}{3} \right) &= \left( \dfrac{1}{3} \right)^3 - \dfrac{1}{2} \left( \dfrac{1}{3} \right)^2 \\[6pt] &= \dfrac{1}{27} - \dfrac{1}{18} \\[6pt] &= -\dfrac{1}{54} \end{align} $$

45.The point $\left( \dfrac{1}{3}, -\dfrac{1}{54} \right)$ is therefore a local minimum.

How to determine the concavity of a function

46.Let $y = f(x)$ be a function that is continuous and twice differentiable in a neighborhood of the point $x_0.$

47.If at $x_0$ we have $f''(x_0) \neq 0,$ then:

49.We return to the function of Example 1 and determine its convexity and concavity:

$$y = f(x) = x^3 - \frac{1}{2}x^2$$

50.The second derivative of the function is:

$$f''(x) = 6x - 1$$

51.We study the sign by imposing:

$$6x - 1 > 0 \implies x > \frac{1}{6}$$

52.The sign chart gives the intervals in which the function is concave upward or concave downward:

54.The function is therefore concave downward for $x < 1/6$ and concave upward for $x > 1/6.$ Since the second derivative changes sign at $x = 1/6,$ the function has an inflection point there. The corresponding ordinate is:

$$f\left( \dfrac{1}{6} \right) = \dfrac{1}{216} - \dfrac{1}{2} \cdot \dfrac{1}{36} = \dfrac{1}{216} - \dfrac{3}{216} = -\dfrac{1}{108}$$

55.Since $f'\left( \dfrac{1}{6} \right) = \dfrac{3}{36} - \dfrac{1}{6} = -\dfrac{1}{12} \neq 0,$ the tangent line at the inflection point is neither horizontal nor vertical, and the point $\left( \dfrac{1}{6}, -\dfrac{1}{108} \right)$ is an oblique inflection point.

Identifying inflection points

56.A point $x_0$ is an inflection point when the second derivative $f''(x)$ changes sign as $x$ crosses $x_0.$ The type of inflection depends on the behavior of the first derivative at $x_0.$