Problem 39 - Essential Calculus Early Transcendentals

Question

(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 \(f .\) \(f(x)=\sqrt{x^{2}+1}-x\)

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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 \( y = 0 \).

Step 3: 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 \( (\frac{1}{\sqrt{2}}, \infty) \).

Step 4: 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 \( 1 \).

Step 5: 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 \( (-\infty, \infty) \).

Step 6: 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.

Key Concepts

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.