Problem 44 - Calculus Early Transcendentals

Question

Suppose that a function \(f\) is continuous on \([0,1]\) except at 0.25 and that \(f(0)=1\) and \(f(1)=3 .\) Let \(N=2 .\) Sketch two possible graphs of \(f\) , one showing that \(f\) might not satisfy the conclusion of the Intermediate Value Theorem and one showing that \(f\) might still satisfy the conclusion of the Intermediate Value Theorem (even though it doesn't satisfy the hypothesis).

Step-by-Step Solution

Answer

Two graphs could show whether \( f \) meets Intermediate Value Theorem's conclusion, depending on continuity around \( N = 2 \).

Step 1: Understanding the Function

The function \( f \) is continuous on the interval \([0, 1]\), except at \( x = 0.25 \), with known values \( f(0) = 1 \) and \( f(1) = 3 \). \( N = 2 \) is within the range of \( f \) from \( f(0) \) to \( f(1) \). Our goal is to illustrate cases where the Intermediate Value Theorem (IVT) might or might not apply.

Step 2: Sketching the Function Not Satisfying IVT

To show a graph where \( f \) does not satisfy the conclusion of the IVT, plot a graph starting at \( (0,1) \), then let the function jump to a value above 2 after \( x = 0.25 \), before decreasing back to \( (1,3) \). Due to the jump at \( x = 0.25 \), there is a gap at \( N = 2 \) which violates the IVT, as the function is not continuous across the interval.

Step 3: Sketching the Function Satisfying IVT

To construct a graph where \( f \) does satisfy the IVT despite the discontinuity, start from \( (0,1) \) and create a curve passing through \( y = 2 \) before reaching \( x = 0.25 \) and continue towards \( (1,3) \). On achieving values that pass through 2, even with a discontinuity at 0.25, the graph exemplifies the conclusion of the IVT: \( f(x) = 2 \) for some \( x \in (0, 0.25) \).

Key Concepts

Function Continuity

Function continuity is a fundamental concept in calculus that describes how a function behaves across different points in its domain. Essentially, a continuous function does not have any sudden jumps, breaks, or holes. This means you should be able to draw the graph of the function without lifting your pen from the paper. A function is continuous at a point if:

• The function value exists at that point.
• The limit of the function as it approaches that point exists.
• The limit equals the function value at that point.

In this case, our function is continuous on the interval \[0, 1\], except at \{0.25\}. Despite this discontinuity, we can look at the overall interval to understand its continuous portions.

Discontinuous Function

A discontinuous function is one that does not meet the criteria for continuity at certain points. Discontinuities can arise in different forms such as jumps, infinite discontinuities, or removable discontinuities. In our exercise, the function has a jump discontinuity at \(x = 0.25\). This type of discontinuity occurs when there is a sudden jump in the function's value at a certain point.

To identify a discontinuity in a problem:

• Check if the function value and limits exist at the specified point.
• Determine if the limit equals the function value; a mismatch indicates a discontinuity.

Even with this discontinuity, it's possible for the function to satisfy the Intermediate Value Theorem (IVT) by achieving specified values at other parts of the interval.

Graph Sketching

Graph sketching is a vital skill in calculus used to understand the behavior of functions visually. It involves plotting key points, noting changes in the function, and connecting these smoothly to anticipate how the function behaves over its domain.

• First, identify key points such as intercepts and known values like \(f(0) = 1\) and \(f(1) = 3\).
• Observe where discontinuities occur, like at \(x = 0.25\), and illustrate behavior both before and after the discontinuity.
• For situations like ours with a specific value \(N = 2\) to consider, adjust the curve to either meet or miss this value, illustrating different scenarios.

This method makes it easier to visualize if the function passes through a certain range, such as between \(f(0)\) to \(f(1)\), highlighting the application of theories like the IVT.

Calculus Problem Solving

Solving calculus problems often involves applying learned concepts like continuity, limits, and the Intermediate Value Theorem. The exercise provided requires using these concepts to determine when the IVT can or cannot be applied.

• Start by defining the problem clearly, such as determining where the function is continuous or not.
• Use graphical representations to hypothesize about the behavior of the function at given intervals.
• Apply concepts like the Intermediate Value Theorem to argue whether certain conditions are met, considering areas both with and without discontinuities.

Through calculus problem solving, students learn to analyze and integrate different aspects of a problem. This deepens understanding of the function's behavior over its domain and enhances their analytical capabilities in more complex scenarios.