/*! 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 12 Find all critical points and the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find all critical points and then use the first derivative test to determine local maxima and minima. Check your answer by graphing. $$f(x)=\frac{x}{x^{2}+1}$$

Short Answer

Expert verified
Critical points are at \( x = -1 \) (local minimum) and \( x = 1 \) (local maximum).

Step by step solution

01

Find the First Derivative

To find critical points, we need the first derivative of the function. Start with the original function \( f(x) = \frac{x}{x^2 + 1} \). Apply the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \) where \( u = x \) and \( v = x^2 + 1 \). The derivatives are \( u' = 1 \) and \( v' = 2x \). Thus, \( f'(x) = \frac{(1)(x^2 + 1) - (x)(2x)}{(x^2 + 1)^2} = \frac{x^2 + 1 - 2x^2}{(x^2 + 1)^2} = \frac{1 - x^2}{(x^2 + 1)^2} \).
02

Find Critical Points

Critical points occur where the first derivative is zero or undefined. The fraction \( \frac{1 - x^2}{(x^2 + 1)^2} \) is undefined when the denominator is zero, which does not occur because \( x^2 + 1 eq 0 \) for all real \( x \). Now, set the numerator equal to zero: \( 1 - x^2 = 0 \), leading to \( x^2 = 1 \). Solve for \( x \), giving \( x = \pm 1 \). These are the critical points.
03

Perform the First Derivative Test

To determine whether each critical point is a local maximum or minimum, evaluate the sign of \( f'(x) \) around the critical points. For \( x = -1 \), pick \( x = -2 \) and \( x = 0 \) for testing. At \( x = -2 \), \( f'(-2) = \frac{1 - (-2)^2}{((-2)^2 + 1)^2} = \frac{1 - 4}{5^2} = \frac{-3}{25} < 0 \), indicating decreasing intervals. At \( x = 0 \), \( f'(0) = \frac{1 - 0}{1} = 1 > 0 \), indicating an increasing interval, suggesting a minimum at \( x = -1 \). For \( x = 1 \), pick \( x = 0 \) and \( x = 2 \). We already found \( f'(0) > 0 \). For \( x = 2 \), \( f'(2) = \frac{1 - 4}{5^2} = \frac{-3}{25} < 0 \), indicating decreasing intervals, suggesting a maximum at \( x = 1 \).
04

Graph the Function

Graph \( f(x) = \frac{x}{x^2 + 1} \) to visually confirm the results. At \( x = -1 \), the graph should show a local minimum, and at \( x = 1 \), it should show a local maximum. The behavior of the function around these points should align with our derivative test observations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

First Derivative Test
The first derivative test is a useful tool to determine whether a critical point is a local maximum, a local minimum, or neither. The process begins by finding the first derivative of a function. This derivative reveals the function's slope at any point, allowing us to identify where the function is increasing or decreasing.

Once you have the derivative, you examine the points where it is zero or undefined, as these are the critical points. To apply the first derivative test:
  • Identify the critical points by setting the derivative equal to zero and solving for the variable.
  • Choose test points on intervals around each critical point to evaluate the sign of the derivative in those intervals.
  • Analyze how the sign of the derivative changes as it passes through each critical point.
If the derivative changes from positive to negative at a critical point, the point is a local maximum. Conversely, if it changes from negative to positive, it indicates a local minimum. If the derivative does not change signs, the critical point is neither a maximum nor minimum.
Quotient Rule
The quotient rule is a fundamental technique in calculus used to differentiate functions presented as a ratio or division of two functions.

For the function \[ f(x) = \frac{u(x)}{v(x)} \]where both \( u(x) \) and \( v(x) \) are differentiable, the derivative using the quotient rule is given by:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}\]

To apply the quotient rule:
  • Differential the numerator function \( u \) resulting in \( u' \).
  • Differential the denominator function \( v \) resulting in \( v' \).
  • Substitute these into the formula \( \frac{u'v - uv'}{v^2} \).
  • Simplify the expression if possible to find the first derivative.
This technique is especially useful in problems requiring analysis of rational functions, such as identifying critical points or understanding the local behavior of the graph.
Local Maxima and Minima
Critical points of a function often indicate potential local maxima or minima, which are important for understanding the function's behavior. A local maximum occurs when a function reaches a peak at a certain point, meaning that the function's values are higher at this point than at any nearby points.

Conversely, a local minimum is where the function attains a trough, with values lower compared to surrounding values. Identifying these points involves the following steps:
  • Find the critical points by setting the first derivative equal to zero.
  • Use the first derivative test to analyze the behavior of the function around these points.
  • Look for changes in the direction of the slope (increasing or decreasing) near each critical point.
Graphing the function can further confirm the existence and nature of these local extrema. By visualizing the function, you can verify your calculations and ensure that each identified maximum or minimum aligns with the expected behavior of the function.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(y=A e^{x}+B e^{-x}\) for any constants \(A, B\) (a) Sketch the graph of the function for (i) \(\quad A=1, B=1\) (ii) \(\quad A=1, B=-1\) (iii) \(A=2, B=1\) (iv) \(A=2, B=-1\) (v) \(A=-2, B=-1 \quad\) (vi) \(A=-2, \quad B=1\) (b) Describe in words the general shape of the graph if \(A\) and \(B\) have the same sign. What effect does the sign of \(A\) have on the graph? (c) Describe in words the general shape of the graph if \(A\) and \(B\) have different signs. What effect does the sign of \(A\) have on the graph? (d) For what values of \(A\) and \(B\) does the function have a local maximum? A local minimum? Justify your answer using derivatives.

Are the statements true of false? Give an explanation for your answer. If the radius of a circle is increasing at a constant rate, then so is the circumference.

Investigate the given two parameter family of functions. Assume that \(a\) and \(b\) are positive. (a) Graph \(f(x)\) using \(b=1\) and three different values for \(a\). (b) Graph \(f(x)\) using \(a=1\) and three different values for \(b\). (c) In the graphs in parts (a) and (b), how do the critical points of \(f\) appear to move as \(a\) increases? As \(b\) increases? (d) Find a formula for the \(x\) -coordinates of the critical point(s) of \(f\) in terms of \(a\) and \(b\). $$f(x)=\frac{a x}{x^{2}+b}$$

A family of functions is given by $$r(x)=\frac{1}{a+(x-b)^{2}}$$ (a) For what values of \(a\) and \(b\) does the graph of \(r\) have a vertical asymptote? Where are the vertical asymptotes in this case? (b) Find values of \(a\) and \(b\) so that the function \(r\) has a local maximum at the point (3,5)

Grit, which is spread on roads in winter, is stored in mounds which are the shape of a cone. As grit is added to the top of a mound at 2 cubic meters per minute, the angle between the slant side of the cone and the vertical remains \(45^{\circ} .\) How fast is the height of the mound increasing when it is half a meter high? [Hint: Volume \(\left.V=\pi r^{2} h / 3, \text { where } r \text { is radius and } h \text { is height. }\right]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.