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Th. Extreme Value. Nested Interval Property.

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Recall

Definition. Let A be a non-empty set of real numbers. A real number S is called an upper bound for A if S ≥ a ∀a ∈ A. A real number S is the least upper bound or supremum for A, and written as S = sup(A), if S is both an upper bound for A and also S ≤ y for every upper bound y of S.

In other words, S is the least upper bound or supremum of a set A ⊆ R if satisfies: (i) S ≥ a ∀a ∈ A, (ii) If u ≥ a ∀a ∈ A ⇒ S ≤ u. Basically, an upper bound is any value larger than or equal to every value of our set A.

Axiom of Completeness

As Abbot put it “we all grow up believing in the existence of real numbers. The property that distinguishes the real numbers is the Axiom of Completeness.”

Axiom of Completeness. Any non-empty set of real numbers that has an upper bound must have a least upper bound or supreme . It is also called the least-upper-bound property.

The result does not hold for the rational numbers ℚ, e.g., S = { x ∈ ℚ: x2 < 2} in $(-\sqrt{2}, \sqrt{2})$. This is true because of the completeness of the real numbers. It implies that there are no “gaps or missing points” in the real number line.

Definitions. The maxima and minima of a function, known collectively as extrema, are the largest and smallest values of the function, either within a given range (the local or relative extrema), or on the entire domain.

Definitions. f(x) has an absolute or global maximum at x = c if f(x) ≤ f(c) ∀x ∈ D, that is, for every x in the domain we are working on. f(x) has an absolute or global minimum at x = c if f(x) ≥ f(c) ∀x ∈ D.

f(x) has a relative global maximum at x = c if f(x) ≤ f(c) ∀x in the neighborhood of c, that is, for every x in (c-ε, c+ε). f(x) has a relative minimum at x = c if f(x) ≥ f(c) ∀x in the neighborhood of c, ∀x ∈ (c-ε, c+ε).

Boundedness theorem. If f(x) is continuous on [a,b] then it is bounded on [a,b].

Nested Interval Property

Nested Interval Property. For each n ∈ ℕ, suppose we have a closed interval In = [an, bn] = {x ∈ ℝ | an ≤ x ≤ bn} such that I1 ⊇ I2 ⊇ I3 ··· ⊇ In ⊇ ···. Then, $\bigcap^∞_{i=1} {I_i} ≠ ∅$

Proof. (Figure iv)

 

Let A and B be two sets defined as follows, A = {an | n ∈ ℕ} ⊆ ℝ, B = {bn | n ∈ ℕ} ⊆ ℝ. Please notice that ∀n ∈ ℕ, bn is an upper bound for A.

⇒[Axiom of Completeness. Every nonempty set of real numbers that is bounded above (below) has a least upper (greatest lower) bound.] A has a least upper bound, a = sup(A), hence ∀n ∈ ℕ, an≤ a ≤ bn ⇒ ∀n ∈ ℕ, a ∈ [an, bn] ⇒ $a ∈ \bigcap^∞{i=1} {I_i} ⇒ \bigcap^∞{i=1} {I_i} ≠ ∅.$

Extreme Value Theorem

Extreme Value Theorem. If f is a continuous function on an interval [a,b], then f has both maximum and minimum values on [a,b]. It states that if a function is continuous on a closed interval, then the function must have a maximum and a minimum on the interval.  

It is easy to see why this theorem holds true and why the requirement of continuity is essential, see Figure ii and 1.f. If a function had a vertical asymptote on [a, b] (Figure ii.b, y = $\frac{1}{x}$ on the interval [0, 5]), then the function gets arbitrary large or negative so there is no biggest or most negative value the function takes on the closed interval.

Lemma. Let S be a non-empty set of real numbers with a supremum M. Then, there is a sequence {sn} that converges to the supremum, that is, {sn}→M.

Proof.

For every N consider M -1N. It is pretty obvious that M -1N < M. M is a supremum, so by definition there is some (aka some element in S better than M -1N) Sn∈ S such that M -1N < sn ≤ M. By the Squeeze theorem, M -1N→M, M→M, then sn →M

Proof (Extreme Value Theorem). If f is a continuous function on an interval [a,b], then f has both maximum and minimum values on [a,b]. It states that if a function is continuous on a closed interval, then the function must have a maximum and a minimum on the interval

Let’s show that the function has a maximum. To prove that the function has a minimum, use -f.

By the boundedness theorem [If f(x) is continuous on [a,b] then it is bounded on [a,b]], there is a C > 0, |f(x)| ≤ C ∀x ∈ [a,b]

Let S = {f(x) : x ∈ [a,b]}. Since f is bounded, so S is bounded. Hence (by the least-upper-bound property, it states thatany non-empty set of real numbers that has an upper bound must have a least upper bound or supreme in real numbers.) S has a least upper bound or supremum M.

By our previous lemma [Let S be a non-empty set of real numbers with a supremum M. Then, there is a sequence {sn} that converges to the supremum, that is, {sn}→M], there is a sequence {yn} in S such that {yn} → M

{yn} =[By definition] {f(xn)} → M, xn ∈ [a,b]

Xn is obviously bounded (between a and b), therefore by the Bolzano–Weierstrass theorem (Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝn has a convergent subsequence), there is a convergent subsequence, {xnk} → x0 ∈ [a,b].

Let’s prove f(x0)=M.

Since {xnk} → x0 and f is continuous ⇒ {f(xnk)} → f(x0). We already know that {yn} = {f(xn)} → M, so any subsequence {xnk} must also convert to M, {f(xnk)} → M, and therefore f(x0) = M.

f(x0) = M and M is the least upper bound or supreme of S = {f(x) : x ∈ [a,b]}, f(x0) = M ≥ f(x) ∀x ∈ [a,b], so M is indeed the maximum of f in [a,b].

How to find Absolute Extrema given a function f on a close interval [a, b]

  1. Verify the function is continuous on [a, b]
  2. Calculate the derivate and determine all its critical values.
  3. Evaluate the function at the previous critical values and the endpoints, namely “a” and “b”.
  4. Obviously, the absolute maximum value and absolute minimum value of f corresponds to the largest and smallest y-values respectively found in the previous step.

Solved exercises

 

f is continuous, so by the Extreme Value Theorem f has an absolute maximum and an absolute minimum.

f’(x) = $1·e^{\frac{x}{2}} +x·e^{\frac{x}{2}}·\frac{1}{2} = e^{\frac{x}{2}} (1 + \frac{x}{2})$, f’(x) = 0 ↭[The exponential function is strictly positive] $1 + \frac{x}{2} = 0 ↭ x = -2$

To find any absolute extrema, we can simply evaluate the function at its critical value (-2) and the endpoints (-3 and 1), and we get f(-2) = $-2e^{\frac{-2}{2}} = \frac{-2}{e}≈ -0.74$, f(-3) = $-3e^{\frac{-3}{2}} ≈ -0.67$, and f(1) = $1·e^{\frac{1}{2}} ≈ 1.65$.

Therefore, f has an absolute minimum at -2 (-2, -0.74), and an absolute maximum at (1, 1.65) on the interval [-3, 1].

f is continuous, so by the Extreme Value Theorem f has an absolute maximum and an absolute minimum.

f’(x) = 3·(x2-1)2·2x = 6x(x2-1)2, f’(x) = 0 ↭ Critical points: x = 0, ±1.

To find any absolute extrema, we can simply evaluate the function at its critical values (0, 1, -1) and the endpoints (-2 and 2), and we get (0, -1), (1, 0), (-1, 0), (2, 27), (-2, 27).

Therefore, f has an absolute minimum at 0 (0, -1), and an absolute maximum at the points (2, 27) and (-2, 27) on the interval [-2, 2].

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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