Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch, as best you can, the graph of a function \(f\) that satisfies all the following conditions. (a) Its domain is the interval \([0,4]\). (b) \(f(0)=f(1)=f(2)=f(3)=f(4)=1\) (c) \(\lim _{x \rightarrow 1} f(x)=2\) (d) \(\lim _{x \rightarrow 2} f(x)=1\) (e) \(\lim _{x \rightarrow 3^{-}} f(x)=2\) (f) \(\lim _{x \rightarrow 3^{+}} f(x)=1\)

Step-by-Step Solution

Answer

The function has jumps at \(x = 1\) and \(x = 3\), and otherwise stays at \(y = 1\).

Step 1: Understand the Domain

The function's domain is the interval \([0,4]\). This means the graph should only exist between the x-values of 0 and 4, inclusive.

Step 2: Identify Points for Interpolation

The function must satisfy that \(f(x) = 1\) at \(x = 0, 1, 2, 3, \text{ and } 4\). Plot these points on the graph: 1. \((0, 1)\)2. \((1, 1)\)3. \((2, 1)\)4. \((3, 1)\)5. \((4, 1)\).

Step 3: Apply the Limit Condition at x = 1

According to condition (c), \( \lim_{x \to 1} f(x) = 2 \). This implies that, as \(x\) approaches 1 from either side, \(f(x)\) should approach 2. The graph can be drawn to have a jump discontinuity at \(x = 1\), where the graph jumps to 2 as \(x\) approaches 1 but returns to 1 exactly at \(x = 1\).

Step 4: Apply the Limit Condition at x = 2

As per condition (d), \( \lim_{x \to 2} f(x) = 1 \). This condition aligns with \(f(2) = 1\), so the graph should smoothly pass through the point \((2,1)\) without any discontinuity.

Step 5: Apply the Limit Conditions at x = 3

For condition (e), \( \lim_{x \to 3^-} f(x) = 2 \), meaning as \(x\) approaches 3 from the left, \(f(x)\) approaches 2. For condition (f), \( \lim_{x \to 3^+} f(x) = 1 \), meaning as \(x\) approaches 3 from the right, \(f(x)\) approaches 1. The graph should have a jump from 2 to 1 at \(x = 3\). Plot this discontinuity at \(x = 3\) as a clear jump.

Step 6: Verify Consistency with Conditions

Ensure the graph meets all given conditions: it should start at \((0,1)\), reach a peak near \((1,2)\) with a corresponding drop to \((1,1)\), then proceed smoothly to \((2,1)\) and jump from \((3,2)\) to \((3,1)\) before leveling off to meet \((4,1)\).

Key Concepts

Domain of a function

The domain of a function primarily dictates where the function is defined. In simpler terms, it tells us the set of possible input values (usually denoted as 'x') for which the function can produce an output (usually denoted as 'f(x)'). For the given exercise, the domain is the closed interval \([0, 4]\). This implies that the function exists only for x-values between and including 0 and 4.

When graphing a function, respecting the domain boundaries is crucial. It means you only draw the graph for the interval specified by the domain. Points outside this domain are not considered since the function values are not defined there.

• For example, in the exercise, any x-values outside the interval \([0, 4]\) are irrelevant. We focus solely on this range for our graph.

Understanding the domain ensures the accuracy of the graph and helps to avoid misinterpretations.

Graph of a function

Graphing a function involves plotting its points on a coordinate plane, where the x-axis represents the domain and the y-axis represents the range of the function. The graph provides a visual representation of the mathematical relationship described by the function.

In this exercise, key points were provided as \(f(0)=f(1)=f(2)=f(3)=f(4)=1\). These coordinates \((0,1), (1,1), (2,1), (3,1), (4,1)\) need to be plotted.

• These points indicate specific values of the function where the output is always 1 regardless of the input from this set.

Visualizing these points and connecting them as per the conditions will give us the required curve for the function. The behavior of the function, such as rise and drop at points influenced by limits (like at x=1 and x=3), will have to follow accurately for a correct depiction.

Discontinuities in functions

Discontinuities occur in functions when there's a sudden jump, break, or hole in the graph. They are points where the function isn't continuous, meaning it doesn't seamlessly connect from one point to another. In this context, understanding the types of discontinuities is essential.

In the given exercise:

• A jump discontinuity is observed at x=1 and x=3. The condition \(\lim _{x \rightarrow 1} f(x)=2\) suggests that approaching x=1 from either side leads to a value of 2, although \(f(1)=1\). This gap at x=1 is a clear jump discontinuity.
• Similarly, approaching x=3 from the left leads to \(\lim_{x \rightarrow 3^-} f(x) = 2\) and from the right \(\lim_{x \rightarrow 3^+} f(x) = 1\). This indicates a shift in value at x=3, hence another jump discontinuity.

Recognizing these discontinuities is key because it influences how we connect points on our graph, often requiring clear gaps or jumps in the plotted curve.

Step-by-step graphing

Graphing a function step-by-step involves a methodical process of plotting points and understanding how limits affect these points. Let's break down this process using the exercise as an example:

• **Identify the domain**: Start by marking the boundary points of the domain (here, 0 and 4) on your graph.
• **Plot given points**: Place each exact point the function must pass through, like \( (0,1), (1,1), (2,1), (3,1), (4,1) \), on your graph.
• **Consider limits**: At points with specified limits like \( \lim_{x \to 1} f(x) = 2 \), sketch the arrow indicating the approach to 2 from both sides despite the actual function value being 1.
• **Add discontinuities**: Highlight gaps clearly at x=1 and x=3, where the limits differ from the function value.
• **Connect the dots thoughtfully**: Use a line or curve to connect the points, bearing in mind how the limits and discontinuities affect the overall shape.

By systematically following these steps, you can ensure that the graph accurately represents the function under the specified conditions.