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Chapter 4: Problem 17

Determine whether the statement is true or false. Explain your answer. If a function \(f\) is continuous on \([a, b],\) then \(f\) has an absolute maximum on \([a, b] .\)

Short Answer

Expert verified
The statement is true; by the Extreme Value Theorem, a continuous function on a closed interval has an absolute maximum.

Step by step solution

01

Understanding the Extreme Value Theorem

The Extreme Value Theorem states that if a function \(f\) is continuous on a closed interval \([a, b],\) then \(f\) must attain both an absolute maximum and an absolute minimum on that interval. This means that within the interval, there exist values such that the function achieves its highest and lowest outputs.
02

Identify the Given Conditions

The problem states that the function \(f\) is continuous on the closed interval \([a, b].\) These conditions precisely match those needed to apply the Extreme Value Theorem.
03

Apply the Theorem

Since the given function \(f\) satisfies the conditions of being continuous on the closed interval \([a, b],\) according to the Extreme Value Theorem, it follows that \(f\) must have an absolute maximum (and minimum) on the interval \([a, b].\) Therefore, the statement is true based on this theorem.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Functions
A continuous function is one where the graph doesn't break or jump. When you trace it, you never lift your pen from the paper. Such functions are predictable and smooth. This smoothness is crucial in mathematics, as it ensures consistency in output values.
  • The function has no sudden changes in direction.
  • Every point within the domain is defined and connected to its neighbors.
  • Examples include lines, curves, and waves, without any sharp edges or holes.

The predictability of continuous functions makes them very useful for analysis and problem-solving. They are the backbone of various theorems, including the Extreme Value Theorem, which guarantees reaching certain values within an interval.
Absolute Maximum
The absolute maximum is the highest point over the entire range of a function. In easier terms, it's like the top of a mountain within the interval considered. When a function is continuous over a closed interval, it achieves this peak.
  • It's not just the highest peak we see immediately, but the overall highest in the range.
  • Absolute maximum is unique within its interval.
  • It could be at a point on the inside or at an endpoint of the interval.
The Extreme Value Theorem ensures that for continuous functions observable on a closed interval, there will always be an absolute maximum. This reliability is fundamental in mathematical proofs and calculations.
Closed Intervals
Closed intervals are segments of the number line that include their endpoints. This means if the interval is denoted as \( [a, b] \), both the numbers \(a\) and \(b\) are included in the set.
  • Closed intervals are represented with square brackets: \[a, b\].
  • They guarantee no loss of boundary values, providing a complete set.
  • It is extended to include the limits, which is crucial for applying theorems like the Extreme Value Theorem.

In contexts like the Extreme Value Theorem, having a closed interval allows us to claim that maximum and minimum values indeed exist. Continuous functions, when applied over closed intervals, promise a reliable examination of real-world scenarios.

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Most popular questions from this chapter

(a) Let $$ f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 0} \\ {x^{2}+1,} & {x>0}\end{array}\right. $$ Show that $$ \lim _{x \rightarrow 0^{-}} f^{\prime}(x)=\lim _{x \rightarrow 0^{+}} f^{\prime}(x) $$ but that \(f^{\prime}(0)\) does not exist. (b) Let $$ f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 0} \\ {x^{3},} & {x>0}\end{array}\right. $$ Show that \(f^{\prime}(0)\) exists but \(f^{\prime \prime}(0)\) does not.

Suppose that \(x=x_{0}\) is a point at which a function \(f\) is continuous but not differentiable and that \(f^{\prime}(x)\) approaches different finite limits as \(x\) approaches \(x_{0}\) from either side. Invent your own term to describe the graph of \(f\) at such a point and discuss the appropriateness of your term.

Let \(f(x)=\tan x\) (a) Show that there is no point \(c\) in the interval \((0, \pi)\) such that \(f^{\prime}(c)=0,\) even though \(f(0)=f(\pi)=0\). (b) Explain why the result in part (a) does not contradict Rolle's Theorem.

Use a graphing utility to determine how many solutions the equation has, and then use Newton鈥檚 Method to approximate the solution that satisfies the stated condition. \(x-\tan x=0 ; \pi / 2

One way of proving that \(f(x) \leq g(x)\) for all \(x\) in a given interval is to show that \(0 \leq g(x)-f(x)\) for all \(x\) in the interval; and one way of proving the latter inequality is to show that the absolute minimum value of \(g(x)-f(x)\) on the interval is nonnegative. Use this idea to prove the inequalities in these exercises. Prove that \(\cos x \geq 1-\left(x^{2} / 2\right)\) for all \(x\) in the interval \([0,2 \pi] .\)
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