Problem 18 - Thomas Calculus

Question

In Exercises \(15-20,\) sketch the graph of each function and determine whether the function has any absolute extreme values on its domain. Explain how your answer is consistent with Theorem \(1 .\) $$ h(x)=\left\\{\begin{array}{ll}{\frac{1}{x},} & {-1 \leq x<0} \\ {\sqrt{x},} & {0 \leq x \leq 4}\end{array}\right. $$

Step-by-Step Solution

Answer

The function has no absolute extrema on its domain due to a vertical asymptote at \( x=0 \).

Step 1: Understand the Function

The function \( h(x) \) is a piecewise function defined as \( \frac{1}{x} \) for \( -1 \leq x < 0 \), and \( \sqrt{x} \) for \( 0 \leq x \leq 4 \). This means we have two different expressions for different intervals of \( x \) in the domain [-1, 4].

Step 2: Analyze \( h(x) = \frac{1}{x} \) for \(-1 \leq x < 0\)

In the interval \( -1 \leq x < 0 \), the function \( h(x) = \frac{1}{x} \) has a vertical asymptote at \( x = 0 \). As \( x \) approaches 0 from the left, \( h(x) \) tends to negative infinity. As \( x \) approaches -1, \( h(x) = -1 \), so the values of the function range from negative infinity to -1.

Step 3: Analyze \( h(x) = \sqrt{x} \) for \(0 \leq x \leq 4\)

In the interval \( 0 \leq x \leq 4 \), the function \( h(x) = \sqrt{x} \) increases from 0 to 2. So, the smallest value occurs at \( x = 0 \) and is 0, and the largest occurs at \( x = 4 \) and is 2.

Step 4: Combine the Analysis

Overall, within its domain \([-1, 4]\), the function \( h(x) \) does not have an absolute maximum or minimum because as \( x \to 0^- \), \( h(x) \) tends to \(-\infty\). However, \( h(x) = \sqrt{x}\) has a local maximum of 2 at \( x=4 \) and a local minimum of 0 at \( x=0 \).

Step 5: Consistency with Theorem 1

Theorem 1 states that a function defined on a closed interval will have an absolute maximum and minimum if it is continuous on that interval. Since \( h(x) \) is not continuous at \( x = 0 \) due to the vertical asymptote, it does not have an absolute extreme value on its entire domain \([-1, 4]\).

Key Concepts

Absolute Extreme Values

When talking about piecewise functions, we need to consider whether they have absolute extreme values, or simply the absolute minimum and maximum. These values represent the highest and lowest points achieved by the function on its domain.

For the given function, which is defined as \( \frac{1}{x} \) for \(-1 \leq x < 0\) and \( \sqrt{x} \) for \(0 \leq x \leq 4\), there is a challenge in determining these values because the behavior of the function varies in different intervals. Let's understand why our function doesn't have absolute extreme values:

• In the piece \( h(x) = \frac{1}{x} \), as \( x \) approaches 0 from the left, the function dips toward \(-\infty\), preventing an absolute minimum in \([-1, 0)\).
• In the piece \( h(x) = \sqrt{x} \), the function rises from 0 to 2 over \( [0, 4] \). Here, 2 appears as a larger value but doesn't qualify as an absolute maximum when considering the entire domain, eyeing \(-\infty\) from the first piece.

This absence of absolute extreme values highlights how discontinuities and open intervals can impact their existence. Since \( h(x) \) isn't continuous across the entire domain of \([-1, 4]\), it disrupts having defined absolute extrema.

Vertical Asymptote

A vertical asymptote is a line that a function approaches but never actually touches or crosses. The presence of a vertical asymptote in a function often indicates that the function will increase or decrease indefinitely as it nears this line.

In the case of our piecewise function \( h(x) \), there is a vertical asymptote at \( x = 0 \). This is crucial to understand because:

• As \( x \) approaches 0 from the left in the section \( h(x) = \frac{1}{x} \), the function values drop towards \(-\infty\).
• This sudden change in behavior implies that the values of the function become unbounded near \( x=0 \).
• The asymptote at \( x = 0 \) is part of the reason why \( h(x) \) is not continuous across its entire domain\([-1, 4]\).

Understanding vertical asymptotes helps in analyzing the behavior of functions, especially when these asymptotes cause breaks in continuity, leading to changes in the function's value trends as seen in this example.

Continuity in Closed Intervals

Continuity is a crucial factor when evaluating functions over specific domains, especially closed intervals. A function is continuous on a closed interval if it is continuous at every point within that interval, and takes defined values at the endpoints.

For the given function \( h(x) \), which is piecewise defined, there is a noticeable breakdown in continuity at \( x = 0 \). Let's delve deeper:

• The function switches from \( \frac{1}{x} \) to \( \sqrt{x} \) exactly at \( x = 0 \), leading to a discontinuity at this junction point.
• Since \( h(x) \) isn't smoothly defined or connected across \( x=0 \), it fails to be continuous over the closed interval \([-1, 4]\).
• This discontinuity directly affects the existence of absolute extreme values as per Theorem 1, which necessitates continuity for a function to possess absolute maximums and minimums on a closed interval.

Ensuring continuity across intervals is key in finding absolute extrema, which is why gaps or jumps disrupt our ability to pinpoint clear extreme values on the entire domain of a function.