Problem 41 - 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 \(f .\) \(f(x)=e^{-x^{2}}\)

Step-by-Step Solution

Answer

(a) Horizontal asymptote at \( y = 0 \). (b) Increasing on \((-\infty, 0)\), decreasing on \((0, \infty)\). (c) Local max at \((0, 1)\). (d) Concave down \((-\sqrt{1/2}, \sqrt{1/2})\), inflection at \(\pm\sqrt{1/2}\).

Step 1: Understanding Asymptotes

Vertical asymptotes occur where the function is undefined. The function \( f(x) = e^{-x^2} \) is defined for all real \( x \), so there are no vertical asymptotes. The function approaches \( 0 \) as \( x \) approaches \( \pm\infty \), indicating a horizontal asymptote at \( y = 0 \).

Step 2: Finding Intervals of Increase or Decrease

To find where the function is increasing or decreasing, calculate the first derivative: \( f'(x) = -2xe^{-x^2} \). Set the derivative equal to zero to find critical points: \( -2xe^{-x^2} = 0 \), which gives \( x = 0 \). Test intervals around the critical point: for \( x < 0 \), \( f'(x) > 0 \), and for \( x > 0 \), \( f'(x) < 0 \). Therefore, the function is increasing on \( (-\infty, 0) \) and decreasing on \( (0, \infty) \).

Step 3: Determining Local Maximum and Minimum

From the behavior of \( f'(x) \), \( f(x) \) has a local maximum at \( x = 0 \). Substituting \( x = 0 \) into the original function, \( f(0) = e^{0} = 1 \), so there is a local maximum at \( (0, 1) \).

Step 4: Finding Intervals of Concavity and Inflection Points

Calculate the second derivative: \( f''(x) = (4x^2 - 2)e^{-x^2} \). Set \( f''(x) = 0 \) to find potential inflection points: \( 4x^2 - 2 = 0 \) gives \( x = \pm \sqrt{\frac{1}{2}} \). Test intervals of \( x \) between these points: for \( |x| < \sqrt{\frac{1}{2}} \), \( f''(x) < 0 \) (concave down), and for \( |x| > \sqrt{\frac{1}{2}} \), \( f''(x) > 0 \) (concave up). Thus, inflection points are at \( x = \pm \sqrt{\frac{1}{2}} \).

Step 5: Sketching the Graph

Use the information gathered: The graph has a horizontal asymptote at \( y = 0 \), increases to a maximum at \( x = 0 \), is concave down between \( x = -\sqrt{\frac{1}{2}} \) and \( x = \sqrt{\frac{1}{2}} \), and has inflection points at \( x = \pm \sqrt{\frac{1}{2}} \). Sketch the graph to reflect increasing behavior to \( x = 0 \), and decreasing thereafter, with indicated concavities and inflection points.

Key Concepts

Asymptotes

When studying the function \( f(x) = e^{-x^2} \), asymptotes are crucial for understanding the behavior of the graph as \( x \) moves toward very large or very small values. Asymptotes can be vertical or horizontal.
Vertical asymptotes occur at values of \( x \) where the function is undefined. However, \( e^{-x^2} \) is defined for all real \( x \), so there are no vertical asymptotes in this case.
Horizontal asymptotes describe the behavior of a function as \( x \) approaches infinity or negative infinity. For \( e^{-x^2} \), as \( x \) becomes very large (positive or negative), \( e^{-x^2} \) tends towards \( 0 \). Thus, \( y = 0 \) is a horizontal asymptote. This means that as you go further left or right along the graph, the y-values tend to "hug" the line \( y = 0 \).
This information is particularly useful when sketching the graph, as it helps predict the structure of the tails of the function.

Critical Points

Critical points are vital to finding areas where a function changes from increasing to decreasing, or vice versa. To find critical points of the function \( f(x) = e^{-x^2} \), we begin by taking the derivative, \( f'(x) = -2x e^{-x^2} \).
Setting the derivative equal to zero and solving for \( x \) gives the critical point at \( x = 0 \). Critical points are points where the function may change direction.

• If \( f'(x) > 0 \), the function is increasing.
• If \( f'(x) < 0 \), the function is decreasing.

By analyzing the derivative, the function increases on the interval \((-\infty, 0)\) and decreases on the interval \((0, \infty)\). This indicates a change in direction at \( x = 0 \), making it a point of local maximum.

Concavity

Concavity describes whether the graph of a function curves upwards or downwards. This is determined using the second derivative. Calculating the second derivative of \( f(x) = e^{-x^2} \) gives \( f''(x) = (4x^2 - 2) e^{-x^2} \).
Set \( f''(x) = 0 \) to find potential inflection points. Solving \( 4x^2 - 2 = 0 \) provides solutions \( x = \pm \sqrt{\frac{1}{2}} \). These are the points where the concavity might change.
To understand these changes:

• For \( |x| < \sqrt{\frac{1}{2}} \), \( f''(x) < 0 \), meaning the graph is concave down.
• For \( |x| > \sqrt{\frac{1}{2}} \), \( f''(x) > 0 \), indicating concave up.

This transition at \( x = \pm \sqrt{\frac{1}{2}} \) identifies these as inflection points, where the graph shifts from concave down to concave up.

Derivatives

Derivatives are foundational tools in calculus used to study the rates of change of functions. The first derivative of a function, \( f'(x) \), provides insight into the slope of the function at any given point. For \( f(x) = e^{-x^2} \), the first derivative \( f'(x) = -2x e^{-x^2} \) helps determine where the function is increasing or decreasing.
The second derivative, \( f''(x) \), reveals how the slope itself is changing, which relates to the concavity of the function. For \( f(x) = e^{-x^2} \), \( f''(x) = (4x^2 - 2) e^{-x^2} \) is crucial in finding inflection points and intervals of concavity.
Understanding derivatives and their roles allows us to plot and predict the behaviors of functions more accurately. They are essential for identifying critical points, finding local maxima or minima, and describing the shape and curvature of graphs.