Problem 33 Sketch the graph of $$ f(x)=... [FREE SOLUTION]

Chapter 2: Problem 33

Sketch the graph of $$ f(x)=\left\\{\begin{aligned} -x & \text { if } x<0 \\ x & \text { if } 0 \leq x<1 \\ 1+x & \text { if } x \geq 1 \end{aligned}\right. $$ Then find each of the following or state that it does not exist. (a) $\lim _{x \rightarrow 0} f(x)$ (b) $\lim _{x \rightarrow 1} f(x)$ (c) $f(1)$ (d) $\lim _{x \rightarrow 1^{+}} f(x)$

Short Answer

Expert verified

(a) 0; (b) Does not exist; (c) 2; (d) 2.

Step by step solution

Understand the Function

The function $f(x)$ is defined in three pieces: 1. For $x < 0$, $f(x) = -x$.2. For $0 \leq x < 1$, $f(x) = x$.3. For $x \geq 1$, $f(x) = 1+x$. These pieces define how the graph will be sketched and used to find limits and evaluate the function.

Sketch the Graph

To sketch the graph, break it down into three parts based on the conditions:1. For $x < 0$, plot $f(x) = -x$, which is a line with a negative slope.2. For $0 \leq x < 1$, plot $f(x) = x$, a line with a positive slope starting at the origin.3. For $x \geq 1$, plot $f(x) = 1+x$, a line that starts at the point (1, 2) since $f(1) = 2$ and has a slope of 1.

Evaluate the Limit as x Approaches 0

(a) Notice that as $x$ approaches 0 from the left, $f(x) = -x$ approaches 0. As $x$ approaches 0 from the right, $f(x) = x$ also approaches 0. Thus, $\lim_{x \to 0} f(x) = 0$.

Evaluate the Limit as x Approaches 1

(b) As $x$ approaches 1 from the left, $f(x) = x$ approaches 1. As $x$ approaches 1 from the right, $f(x) = 1+x$ approaches 2. Since the left and right limits are not equal, $\lim_{x \to 1} f(x)$ does not exist.

Evaluate the Function at x=1

Evaluate the Right-Hand Limit at x=1

(d) For the right-hand limit, consider $x > 1$. In this case, $f(x) = 1+x$, and as $x$ approaches 1 from the right, $f(x)$ approaches $2$. Thus, $\lim_{x \to 1^+} f(x) = 2$.

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Key Concepts

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

graph sketching

Graph sketching is an important step in understanding and solving problems involving functions, especially piecewise ones. It involves visually representing how a function behaves across different intervals. For the function given in the exercise, we sketch it by considering each piece separately:

For the interval where $x < 0$, we have $f(x) = -x$. This part of the graph is a straight line with a negative slope, taking values that are reflections of positive numbers in the negative half of the x-axis.
In the section where $0 \leq x < 1$, the function becomes $f(x) = x$. Here, the graph is a diagonal line with a positive slope passing through the origin. This indicates a direct relationship where the function value equals the x-coordinate, resulting in a line that is at 45 degrees to the axes.
Lastly, for $x \geq 1$, the function changes to $f(x) = 1 + x$. This line starts at the point (1, 2) due to the inclusion of 1 in the piece, and it continues with a positive slope. This shift upwards by 1 unit from the function $f(x) = x$ suggests a y-intercept that is higher than the previous segments.

Sketching the graph helps in visualizing discontinuities, intercepts, and slope changes, serving as a guide to tackle further analyses, such as evaluating limits.

limits and continuity

Understanding limits and continuity is key when dealing with piecewise functions. A limit evaluates what value a function approaches as the input gets close to a particular point. For piecewise functions, we often look at one-sided limits to address different behavior in different sections:

At $x = 0$, we consider limits from both sides. From the left $(x \to 0^-)$, $f(x) = -x$ approaches 0. From the right $(x \to 0^+)$, $f(x) = x$ also approaches 0. Therefore, the two sides agree, and the limit $\lim_{x \to 0} f(x) = 0$.
However, at $x = 1$, things change. From the left $(x \to 1^-)$, $f(x) = x$ approaches 1. From the right $(x \to 1^+)$, $f(x) = 1 + x$ approaches 2. The left and right limits differ, indicating a discontinuity, and thus $\lim_{x \to 1} f(x)$ does not exist.

Continuity requires that these limits be equal to the function's actual value at a point, but here $f(x)$ is not continuous at $x = 1$. Understanding these disparities is crucial for predicting function behavior and solving calculus problems effectively.

calculus problem solving

Calculus problem solving often leverages the concepts of limits, continuity, and graph analysis to find solutions to function-based queries. In this exercise, we've navigated various concepts to arrive at clear solutions:

By understanding the function's definition, we determined the behavior in distinct areas and sketched these accordingly.
We evaluated limits at specific points, allowing us to examine whether the function's behavior was consistent (i.e., continuous) or if there were discontinuities.
Finally, by calculating particular values like $f(1)$, we applied these limits and sketches to validate our understanding of the function's characteristics and properly solve queries regarding its behavior around critical points.

These steps are crucial. They offer a deep understanding of the piecewise function and ensure that even complex function rules become manageable. Mastering the breakdown and analysis of such problems is essential in calculus to ensure proficient problem-solving abilities.

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Short Answer

Step by step solution

Understand the Function

Sketch the Graph

Evaluate the Limit as x Approaches 0

Evaluate the Limit as x Approaches 1

Evaluate the Function at x=1

Evaluate the Right-Hand Limit at x=1

Key Concepts

graph sketching

limits and continuity

calculus problem solving

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