/*! 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 23 Suppose \(f "\) is continuous on... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 23

Suppose \(f "\) is continuous on \((-\infty, \infty).\) $$\begin{array}{l}{\text { (a) If } f^{\prime}(2)=0 \text { and } f^{\prime \prime}(2)=-5, \text { what can you say about f? }} \\ {\text { (b) If } f^{\prime}(6)=0 \text { and } f^{\prime \prime}(6)=0 \text { , what can you say about } f ?}\end{array}$$

Short Answer

Expert verified
At \( x = 2 \), \( f \) has a local maximum; at \( x = 6 \), the nature is inconclusive.

Step by step solution

01

Interpreting First and Second Derivative

When \( f'(x) = 0 \), it means that the function \( f(x) \) has a critical point at that \( x \)-value. The second derivative \( f''(x) \) helps us determine the nature of this critical point. If \( f''(x) < 0 \), the critical point is a local maximum; if \( f''(x) > 0 \), it is a local minimum; and if \( f''(x) = 0 \), the test is inconclusive.
02

Analyze Point at \( x = 2 \)

Given \( f'(2) = 0 \) and \( f''(2) = -5 \), since \( f''(2) < 0 \), the function \( f \) has a local maximum at \( x = 2 \). This means the function reaches at least a temporary peak at this point.
03

Analyze Point at \( x = 6 \)

Given \( f'(6) = 0 \) and \( f''(6) = 0 \), the second derivative test is inconclusive. This critical point could be a minimum, maximum, or an inflection point. Further analysis, possibly looking at higher derivatives or graph behavior, would be necessary to determine its exact nature.

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Key Concepts

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

Critical Points
A critical point of a function occurs where the function's derivative is zero, or where the derivative does not exist. In mathematical terms, for a function \( f(x) \), a critical point at \( x = c \) means \( f'(c) = 0 \) or \( f' \) does not exist. These points are crucial in calculus because they potentially indicate where the function has a relative maximum or minimum.
  • Finding critical points is the first step in analyzing the behavior of a function's graph.
  • Checking where the derivative equals zero helps locate points where the slope of the tangent is horizontal.
  • If a function is continuous and differs smoothly, these points can show where the function could have peaks or troughs or even changes in concavity.
For instance, if a problem states that \( f'(2) = 0 \), then \( x = 2 \) is a critical point for the function. This indicates that something noteworthy is happening in the behavior of the function at that point.
First Derivative Test
The first derivative test is a method to determine whether a critical point is a local maximum, minimum, or neither. This test involves looking at the sign of the derivative before and after the critical point. Here's how it works:
  • If \( f'(x) \) changes from positive to negative at \( x = c \), then \( f(c) \) is a local maximum.
  • If \( f'(x) \) changes from negative to positive at \( x = c \), then \( f(c) \) is a local minimum.
  • If \( f'(x) \) does not change sign, then \( f(c) \) is neither a maximum nor a minimum, perhaps a point of inflection instead.
This method is valuable as it provides a simple way to classify critical points just based on the job of the function's slope, offering immediate insights into the function's behavior without requiring examination of higher order derivatives.
Second Derivative Test
The second derivative test offers further insights into the nature of a critical point by using the second derivative of the function. If you have a critical point where the first derivative is zero, you might apply this test to determine its nature:
  • When \( f''(x) > 0 \) at a critical point, the test indicates a local minimum because the function is curving upwards at that point.
  • When \( f''(x) < 0 \), the test indicates a local maximum; here, the graph of the function is curving downwards.
  • If \( f''(x) = 0 \), the test is inconclusive; the point could be a local extremum or an inflection point.
This test is extremely efficient because it allows a function's concavity to signal the presence of a maximum or minimum, assisting in the analysis of function behavior around critical points.
Concavity
Concavity describes the direction a curve opens. Understanding concavity helps describe how a function is behaving as you move along its graph. A function can be:
  • Concave Up: If \( f''(x) > 0 \), the graph of the function is concave up, resembling the shape of a cup (\( \smile \)). Here, the slope of the tangent increases as \( x \) increases.
  • Concave Down: If \( f''(x) < 0 \), the graph is concave down, resembling an upside-down cup (\( \frown \)). The slope of the tangent decreases as \( x \) increases.
Recognizing concavity is vital to understanding the overall shape and characteristics of a function's graph. Moreover, it elucidates how dramatically a function's rate of change is altering at any given interval, which is important for predicting future behavior and values.
Inflection Points
An inflection point occurs at a point where the function changes concavity from up to down or vice versa. These points can be identified on a graph where the curve shifts its direction of bend.
  • Inflection points happen where the second derivative shifts sign.
  • If \( f''(c) = 0 \) or \( f''(c) \) does not exist, and the concavity changes around \( x = c \), \( x = c \) is an inflection point.
  • Graphically, this is where the curve departs from looking like \( \smile \) to \( \frown \), or the other way around.
Inflection points are key to fully understanding a graph because they mark more subtle changes that occur beyond local maxima and minima, revealing where acceleration or deceleration in the graph's growth or decline is most pronounced.

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Most popular questions from this chapter

\(5-8\) Use Newton's method with the specified initial approximation \(x_{1}\) to find \(x_{3},\) the third approximation to the root of the given equation. (Give your answer to four decimal places.) \(x^{5}+2=0, \quad x_{1}=-1\)

\(23-28\) Use Newton's method to find all the roots of the equation correct to eight decimal places. Start by drawing a graph to find initial approximations. \(x^{6}-x^{5}-6 x^{4}-x^{2}+x+10=0\)

The first appearance in print of I'Hospital's Rule was in the book Analyse des Infiniment Petits published by the Marquis de I'Hospital in \(1696 .\) This was the first calculus textbook ever published and the example that the Marquis used in that book to illustrate his rule was to find the limit of the function $$y=\frac{\sqrt{2 a^{3} x-x ^ {4}}-a \sqrt[3]{\operatorname{aax}}}{a-\sqrt[4]{a x^{3}}}$$ as \(x\) approaches a, where a \( > 0 .\) (At that time it was common to write aa instead of a'.) Solve this problem.

Investigate the family of curves given by the equation \(f(x)=x^{4}+c x^{2}+x .\) Start by determining the transitional value of \(c\) at which the number of inflection points changes. Then graph several members of the family to see what shapes are possible. There is another transitional value of \(\mathrm{c}\) at which the number of critical numbers changes. Try to discover it graphically. Then prove what you have discovered.

The linear density of a rod of length 1 \(\mathrm{m}\) is given by \(\rho(\mathrm{x})=1 / \sqrt{\mathrm{x}},\) in grams per centimeter, where \(\mathrm{x}\) is measured in centimeters from one end of the rod. Find the mass of the rod.
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