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Chapter 4: Problem 39

(a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts ( d ) to sketch the graph of \(f .\) \(f(x)=\sqrt{x^{2}+1}-x\)

Short Answer

Expert verified
No vertical asymptotes, horizontal asymptote is y = 0. Increasing on (-1/鈭�2, 1/鈭�2), decreasing elsewhere. Local max at x = 0. Concave up on entire domain. No inflection points.

Step by step solution

01

Identify Vertical Asymptotes

To find the vertical asymptotes of the function, we need to examine where the function is undefined. The function \( f(x) = \sqrt{x^2 + 1} - x \) is defined for all real numbers because the expression inside the square root, \( x^2 + 1 \), is always positive and never zero. Thus, there are no vertical asymptotes.
02

Identify Horizontal Asymptotes

Check the behavior of the function as \( x \to \pm\infty \). Dividing both the numerator and the denominator by \( x \), consider \[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right) = \lim_{x \to \infty} \frac{x^2 + 1 - x^2}{\sqrt{x^2 + 1} + x} = 0 \]. Thus, the horizontal asymptote is \( y = 0 \).
03

Find Intervals of Increase or Decrease

To find intervals where the function is increasing or decreasing, compute the first derivative. \[ f'(x) = \frac{x}{\sqrt{x^2 + 1}} - 1 \]. Set \( f'(x) = 0 \) to find critical points: \[ \frac{x}{\sqrt{x^2 + 1}} - 1 = 0 \Rightarrow x = \pm \frac{1}{\sqrt{2}} \]. Check sign changes of \( f'(x) \) in intervals: \(( -\infty, -\frac{1}{\sqrt{2}} )\), \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\), and \( (\frac{1}{\sqrt{2}}, \infty) \). The function increases on \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\) and decreases on \(( -\infty, -\frac{1}{\sqrt{2}} )\) and \( (\frac{1}{\sqrt{2}}, \infty) \).
04

Determine Local Maximum and Minimum Values

Using the critical points found in Step 3, we identify that there is a local maximum at \( x = 0 \) because \( f'(x) \) changes from positive to negative. Evaluating \( f(0) \) gives \( \sqrt{1} = 1 \), hence local maximum value is \( 1 \).
05

Find Intervals of Concavity

Calculate the second derivative to find concavity: \[ f''(x) = \frac{1}{(x^2 + 1)^{3/2}} \]. Since \( f''(x) > 0 \) for all \( x \), the function is concave up over the entire domain \( (-\infty, \infty) \).
06

Identify Inflection Points

The absence of sign change in \( f''(x) \) implies no inflection points. The function's concavity does not change across any interval.

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Key Concepts

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

Vertical and Horizontal Asymptotes
Asymptotes are lines that the graph of a function approaches but never actually touches. There are two types we often look for: vertical and horizontal.
  • **Vertical Asymptotes:** These occur where a function is undefined, such as points where a denominator might be zero. For our function, \( f(x) = \sqrt{x^2 + 1} - x \), it's defined for all \( x \), so there are no vertical asymptotes.

  • **Horizontal Asymptotes:** These describe the end behavior of a graph as \( x \to \infty \) or \( x \to -\infty \). We found that as \( x \to \infty \), \( f(x) \) approaches 0, giving us a horizontal asymptote at \( y = 0 \). This tells us that far away from the origin, the graph flattens out along this line.
Critical Points and Intervals of Increase or Decrease
Critical points occur where the derivative of a function is zero or undefined; they are potential locations for local maxima or minima. To determine these for \( f(x) = \sqrt{x^2 + 1} - x \), we computed the first derivative, \( f'(x) = \frac{x}{\sqrt{x^2 + 1}} - 1 \).
  • **Critical Points:** Setting \( f'(x) = 0 \), we find critical points at \( x = \pm \frac{1}{\sqrt{2}} \).

  • **Intervals of Increase or Decrease:** By checking the sign of \( f'(x) \):
    • The function increases on the interval \((-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\).

    • It decreases on \(( -\infty, -\frac{1}{\sqrt{2}} )\) and \( (\frac{1}{\sqrt{2}}, \infty) \). This pattern shows a peak at the origin, guiding the overall shape of our graph.
Local Maximum and Minimum Values
Local maxima and minima are found at critical points where the function shifts from increasing to decreasing or vice versa.
  • **Local Maximum:** At a critical point \( x = 0 \), \( f'(x) \) changes from positive to negative, indicating a local maximum. Evaluating the function at this point gives \( f(0) = 1 \).

  • **Local Minimum:** The function has no local minimum since there are no other critical points where \( f'(x) \) changes from negative to positive.
Understanding these values is crucial as they help sketch the general outline of the graph, indicating peaks or troughs.
Concavity and Inflection Points
Concavity refers to the direction the graph curves, and inflection points are where it changes concavity.
  • **Concavity:** To assess this, we calculated the second derivative: \( f''(x) = \frac{1}{(x^2 + 1)^{3/2}} \). As \( f''(x) > 0 \) for all values of \( x \), the function is concave up everywhere. This means the graph always forms a 'U' shape.

  • **Inflection Points:** These are points where the concavity changes. Since \( f''(x) \) remains positive, there are no inflection points, keeping the graph's curvature consistent.
This information is critical in graph sketching and predicting how a curve will behave between and beyond critical points.

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Most popular questions from this chapter

A car is traveling at 50 \(\mathrm{mi} / \mathrm{h}\) when the brakes are fully applied, producing a constant deceleration of 22 \(\mathrm{ft} / \mathrm{s}^{2} .\) What is the distance traveled before the car comes to a stop?

A rain gutter is to be constructed from a metal sheet of width 30 \(\mathrm{cm}\) by bending up one-third of the sheet on each side through an angle \(\theta .\) How should \(\theta\) be chosen so that the gutter will carry the maximum amount of water?

\(49-52=\) The line \(y=m x+b\) is called a slant asymptote if \(f(x)-(m x+b) \rightarrow 0\) as \(x \rightarrow \infty\) or \(x \rightarrow-\infty\) because the vertical distance between the curve \(y=f(x)\) and the line \(y=m x+b\) approaches 0 as \(x\) becomes large. Find an equation of the slant asymptote of the function and use it to help sketch the graph. [ For rational functions, a slant asymptote occurs when the degree of the numerator is one more than the degree of the denominator. To find it, use long division to write $$f(x)=m x+b+R(x) / Q(x) ]$$ $$y=\frac{1+5 x-2 x^{2}}{x-2}$$

Describe how the graph of \(f\) varies as \(c\) varies. Graph several members of the family to illustrate the trends that you discover. In particular, you should investigate how maximum and minimum points and inflection points move when \(c\) changes. You should also identify any transitional values of \(c\) at which the basic shape of the curve changes. $$f(x)=\sqrt{x^{4}+c x^{2}}$$

\(23-36=\) Find the critical numbers of the function. $$f(x)=x^{2} e^{-3 x}$$
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