/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Sketch the graph of the given fu... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the graph of the given function, indicating (a) \(x\) - and \(y\) -intercepts, (b) extrema, (c) points of inflection, \((d)\) behavior near points where the function is not defined, and (e) behavior at infinity. Where indicated, technology should be used to approximate the intercepts, coordinates of extrema, and/or points of inflection to one decimal place. Check your sketch using technology. \(f(x)=x+\frac{1}{x}\)

Short Answer

Expert verified
In summary, the function \(f(x)=x+\frac{1}{x}\) has no x- and y-intercepts, a local minimum at (1,2), a local maximum at (-1,-2), points of inflection at x=0 (undefined), a slant asymptote at y=x as x approaches \(\pm\infty\), and undefined behavior at x=0. The graph can be sketched using this information, and verified with technology for accuracy.

Step by step solution

01

Find the first and second derivatives of f(x)

Using the power rule, calculate the first derivative of the function \(f(x)\): \(f'(x) = \frac{d}{dx}(x+\frac{1}{x}) = 1 - \frac{1}{x^2}\) Now, find the second derivative of the function \(f(x)\): \(f''(x) = \frac{d^2}{dx^2}(1 - \frac{1}{x^2}) = \frac{2}{x^3}\)
02

Find the x- and y- intercepts

To find the x-intercepts, set f(x) equal to 0 and solve for x: \(x+\frac{1}{x}=0\) \(x^2+1=0\) \(x^2=-1\) There are no real solutions, so there are no x-intercepts. To find the y-intercept, set x=0 in the function: \(f(0) = 0 + \frac{1}{0}\) However, the function is not defined when x=0. So, there is no y-intercept.
03

Find the critical points

To find the critical points, set the first derivative of f(x) equal to 0 and solve for x: \(f'(x) = 1 - \frac{1}{x^2} =0\) \(\frac{1}{x^2}=1\) \(x^2=1\) \(x=\pm 1\) So, there are critical points at x=1 and x=-1.
04

Determine the extrema

Since f''(x) = \(\frac{2}{x^3}\), the second derivative test can be applied to determine whether the critical points are local maxima or minima. Evaluate the second derivative at each critical point: For x = 1: \(f''(1) = \frac{2}{1^3} = 2\), which is positive, indicating that there is a local minimum at x = 1. Now, find the corresponding y-coordinate: \(f(1)= 1+\frac{1}{1}=2\) For x = -1: \(f''(-1) = \frac{2}{(-1)^3} = -2\), which is negative, indicating that there is a local maximum at x = -1. Find the corresponding y-coordinate: \(f(-1)= -1+\frac{1}{-1}=-2\) Thus, there is a local minimum at (1,2) and a local maximum at (-1,-2).
05

Identify points of inflection

A point of inflection occurs where the second derivative changes its sign. Since \(f''(x) = \frac{2}{x^3}\), the sign of the second derivative only changes when x changes sign. Thus, there are points of inflection at x = 0. However, the function is not defined at x = 0, so we can't find the corresponding y-coordinate.
06

Analyze behavior at infinity and undefined points

As x approaches positive infinity, the term \(\frac{1}{x}\) approaches 0, so f(x) will approach x, indicating a slant asymptote at y = x. Similarly, as x approaches negative infinity, the term \(\frac{1}{x}\) approaches 0, so f(x) will approach x, indicating a slant asymptote at y = x. As x approaches 0, the term \(\frac{1}{x}\) becomes increasingly large and positively/negatively infinite, depending on the sign of x.
07

Sketch the graph

Using the information from Steps 1-6, sketch the graph of the function, considering the following aspects: - No x- and y-intercepts - Local minimum at (1,2) and local maximum at (-1,-2) - Points of inflection at x = 0 (undefined) - Behavior at infinity (slant asymptote at y = x) - Behavior near points where the function is not defined (x = 0) Now, verify the sketch using technology to ensure the accuracy of the graph.

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Key Concepts

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

First and Second Derivatives
Derivatives are crucial when analyzing functions in calculus. They help us understand how a function changes at any given point. The first derivative, denoted as \( f'(x) \), gives us the slope of the tangent line to the function at a particular point. For the function \( f(x) = x + \frac{1}{x} \), the first derivative is \( f'(x) = 1 - \frac{1}{x^2} \). This indicates how steep or shallow the function is at different values of \( x \).

The second derivative, \( f''(x) \), tells us about the curvature of the function. It reveals where the graph is concave up or concave down. For our function, \( f''(x) = \frac{2}{x^3} \). A positive second derivative means the graph is concave up (like a cup), while a negative second derivative means it's concave down (like a frown). Understanding these derivatives is vital in graph sketching as they inform us about the shape and turning points of the graph.
Critical Points
Critical points of a function occur where its first derivative is zero or undefined. These points are significant as they often indicate where the graph has local maxima or minima. To find these points for \( f(x) = x + \frac{1}{x} \), we solve \( f'(x) = 0 \).

After calculations, we find critical points at \( x = 1 \) and \( x = -1 \). To determine if these are points of local maxima or minima, we use the second derivative test. At \( x = 1 \), since \( f''(1) = 2 \) is positive, the graph is concave up, indicating a local minimum. Conversely, at \( x = -1 \), \( f''(-1) = -2 \) is negative, showing a local maximum. Recognizing these critical points is essential for accurately sketching the function's graph.
Graph Sketching
Graph sketching involves drawing a curve based on calculated key features such as intercepts, extrema, and points of inflection. For the function \( f(x) = x + \frac{1}{x} \), first, observe that there are no real x- or y-intercepts as calculated from \( f(x) = 0 \). Next, we identify the local minimum at \((1, 2)\) and the local maximum at \((-1, -2)\) using the critical point analysis.

Additionally, knowing that there is a point of inflection at \( x = 0 \) helps shape the graph despite the undefined nature at \( x = 0 \). This point indicates a transition in the concavity of the graph. By combining information from derivatives and critical points, we get an accurate layout of the graph, enabling a better understanding of the function's behavior.
Asymptotic Behavior
Asymptotic behavior describes how a function behaves as it approaches infinity or undefined points. For \( f(x) = x + \frac{1}{x} \), understanding this behavior is crucial for graph sketching. As \( x \to \infty \) or \( x \to -\infty \), the term \( \frac{1}{x} \) approaches 0. Therefore, \( f(x) \) behaves like \( x \), indicating a slant asymptote at the line \( y = x \).

Furthermore, as \( x \) nears 0, \( \frac{1}{x} \) grows infinitely large; this is true both positively and negatively, depending on whether \( x \) approaches from the right or left, respectively. These asymptotes and behaviors give us insight into the end behavior of the function, helping in drawing an even more accurate graph representation and understanding where the function might not be defined.

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Most popular questions from this chapter

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