/*! 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 22 Sketch the graph of a function t... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 22

Sketch the graph of a function that satisfies all of the given conditions. \(f^{\prime}(x)>0\) if \(|x|<2, \quad f^{\prime}(x)<0\) if \(|x|>2,\) \(f^{\prime}(2)=0, \quad \lim _{x \rightarrow \infty} f(x)=1, \quad f(-x)=-f(x),\) \(f^{\prime \prime}(x) <0\) if \(0 < x < 3, \quad f^{\prime \prime}(x) > 0\) if \(x > 3\)

Short Answer

Expert verified
Graph is symmetric, increasing around 0, peaks at 2, decreases to 1 as \(x\to \infty\).

Step by step solution

01

Analyze First Derivative Conditions

The condition \( f'(x) > 0 \) when \( |x| < 2 \) tells us that the function is increasing in the interval \((-2, 2)\). The condition \( f'(x) < 0 \) when \( |x| > 2 \) tells us that the function is decreasing when \( x > 2 \) and \( x < -2 \). The point \( f'(2) = 0 \) indicates a critical point at \( x = 2 \), which might be a local maximum or minimum.
02

Analyze Symmetry and Limits

Given \( f(-x) = -f(x) \), the function is odd, meaning it is symmetric with respect to the origin. Also, \( \lim_{x \to \infty} f(x) = 1 \) suggests that the function approaches \( y = 1 \) as \( x \) goes to infinity. This will help us determine the end behavior of the function.
03

Second Derivative Analysis

The condition \( f''(x) < 0 \) for \( 0 < x < 3 \) means the function is concave down on this interval, suggesting a local maximum might occur. Conversely, \( f''(x) > 0 \) for \( x > 3 \) indicates the function is concave up, suggesting a point of inflection at \( x = 3 \).
04

Sketch the Function

Start from the critical point at \( x = 2 \) (local maximum due to increase then decrease), move through the interval \( (-2, 2) \) where the function is increasing. For \( |x| > 2 \), draw the function decreasing. Note the symmetry and that it eventually levels to \( y = 1 \) as \( x \to \infty \) and \( x \to -\infty \). Reflect any features from \( x > 0 \) to \( x < 0 \) using the odd symmetry property.

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Key Concepts

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

First Derivative Test
The first derivative test helps us understand how the function behaves in terms of increasing and decreasing intervals, as well as identifying potential extreme points like local maxima or minima. In this exercise, the first derivative, denoted as \( f'(x) \), changes signs at critical points. Here, you observe the following key conditions:
  • \( f'(x) > 0 \) when \( |x| < 2 \): The function is increasing in this region, meaning it slopes upwards as you move from left to right.
  • \( f'(x) < 0 \) when \( |x| > 2 \): The function is decreasing in these regions, indicating a downward slope.
  • \( f'(2) = 0 \): This indicates a critical point at \( x = 2 \). It implies a potential transition from increasing to decreasing, making it a candidate for a local maximum.
Understanding these changes helps you identify the behavior of the graph and predict its shape around these critical areas.
Second Derivative Test
The second derivative test focuses on the concavity of the function, giving us insights into the curvature and potential points of inflection. Concavity and inflection points provide a deeper layer of understanding about the features of the graph.
  • \( f''(x) < 0 \) for \( 0 < x < 3 \): The function is concave downwards in this interval, suggesting that the graph resembles an upside-down bowl or arc.
  • \( f''(x) > 0 \) for \( x > 3 \): The function is concave upwards, like a U-shape or a smile. This section of the graph opens upwards.
These insights indicate that there is a local maximum around \( x = 2 \) and possibly an inflection point at \( x = 3 \), where the concavity shifts from downward to upward.
Symmetry in Functions
Symmetry in functions simplifies understanding the overall shape of the graph and can dictate significant features, especially around the origin. An odd function, like the one in this exercise, exhibits a particular type of symmetry.
  • \( f(-x) = -f(x) \): This equation points to the function being odd, meaning it has rotational symmetry with respect to the origin. If you graph the function on one side of the \( y \)-axis, the other side is a mirror image, albeit upside-down.
This property helps in sketching the graph, as features from positive \( x \) values can be reflected to negative \( x \) values with the same size but opposite signs.
Critical Points
Critical points are special values of \( x \) where the function's behavior changes, such as from increasing to decreasing, leading to local maxima or minima. These points are crucial in graphing functions and understanding their key features.
  • At \( x = 2 \), the critical point, the first derivative is zero \( f'(2) = 0 \), suggesting a possible local extremum.
Given that the first derivative changes from positive to negative around this point, it indicates a local maximum due to the function increasing and then decreasing. Recognizing critical points can help you predict where peaks or troughs occur in the graph.
Limits at Infinity
The concept of limits at infinity deals with how functions behave as \( x \) approaches very large positive or negative values. Understanding this helps us predict the end behavior or horizontal asymptotes of the function.
  • \( \lim_{x \to \infty} f(x) = 1 \): As \( x \) grows larger and larger, the function approaches the value \( y = 1 \).
This indicates that the graph levels off and does not continue to rise or fall infinitely. Such understanding is essential when considering how to draw the tail ends of the graph and to know that the function stabilizes rather than diverging. This behavior is reflected on both ends of the graph.

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Most popular questions from this chapter

\(53-56\) (a) Use a graph to estimate the absolute maximum and minimum values of the function to two decimal places. (b) Use calculus to find the exact maximum and minimum values. $$f(x)=e^{x}+e^{-2 x}, \quad 0 \leqslant x \leqslant 1$$

A particle is moving with the given data. Find the position of the particle. $$v(t)=1.5 \sqrt{t}, \quad s(4)=10$$

Find \(f\) $$f^{\prime}(x)=5 x^{4}-3 x^{2}+4, \quad f(-1)=2$$

Show that the curve \(y=\sqrt{x^{2}+4 x}\) has two slant asymptotes: \(y=x+2\) and \(y=-x-2 .\) Use this fact to help sketch the curve.

\(37-50=\) Find the absolute maximum and absolute minimum values of \(f\) on the given interval. $$f(t)=2 \cos t+\sin 2 t, \quad[0, \pi / 2]$$
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