Problem 39 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\frac{x^{2}}{1+x^{2}}, x \in \mathbf{R} $$ (a) Determine where \(f(x)\) is increasing and where it is decreasing. (b) Where is the function concave up and where is it concave down? Find all inflection points of \(f(x)\). (c) Find \(\lim _{x \rightarrow \pm \infty} f(x)\) and decide whether \(f(x)\) has a horizontal asymptote. (d) Sketch the graph of \(f(x)\) together with its asymptotes and inflection points (if they exist).

Step-by-Step Solution

Answer

(a) Increasing: \((0, \infty)\), Decreasing: \((-\infty, 0)\). (b) Concave up: \((-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}})\), Concave down elsewhere. Inflection points at \( x = \pm\sqrt{\frac{1}{3}} \). (c) Horizontal asymptote: \( y = 1 \). (d) Graph based on behavior.

Step 1: Finding the First Derivative

The first step to determine where the function \(f(x)\) is increasing or decreasing is to find the first derivative. Given \(f(x) = \frac{x^2}{1+x^2} \), we use the quotient rule: \( f'(x) = \frac{(1+x^2)(2x) - x^2(2x)}{(1+x^2)^2} = \frac{2x + 2x^3 - 2x^3}{(1+x^2)^2} = \frac{2x}{(1+x^2)^2} \).

Step 2: Critical Points and Sign of the First Derivative

Set the first derivative to zero to find critical points: \( f'(x) = 0 \rightarrow \frac{2x}{(1+x^2)^2} = 0 \rightarrow x = 0 \). To determine the sign of \( f'(x) \), observe that \(f'(x)\) is positive for \( x > 0 \) and negative for \( x < 0 \). Thus, \(f(x)\) is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\).

Step 3: Finding the Second Derivative

Calculate the second derivative to determine concavity. The derivative of \( f'(x) = \frac{2x}{(1+x^2)^2} \) using the quotient rule is \( f''(x) = \frac{(1+x^2)^2(2) - 2x(4x)(1+x^2)}{(1+x^2)^4} \). Simplifying gives \( f''(x) = \frac{2 - 6x^2}{(1+x^2)^3} \).

Step 4: Analyzing Concavity

Find where \( f''(x) = 0 \): \( 2 - 6x^2 = 0 \rightarrow x = \pm\sqrt{\frac{1}{3}} \). The sign of \( f''(x) \) indicates concavity: \(f''(x) > 0\) on \(\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) and \(f''(x) < 0\) elsewhere. So, the function is concave up on \(\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) and concave down on \((-\infty, -\frac{1}{\sqrt{3}}) \cup (\frac{1}{\sqrt{3}}, \infty)\). The points \( x = \pm\sqrt{\frac{1}{3}} \) are inflection points.

Step 5: Calculating Limits at Infinity

Find the horizontal asymptotes by calculating the limits: \( \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{x^2}{1+x^2} = 1 \) and \( \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x^2}{1+x^2} = 1 \). Thus, \( y = 1 \) is a horizontal asymptote.

Step 6: Sketching the Graph

Combine all information to sketch the graph. The function is increasing on \((0, \infty)\) and decreasing on \((-\infty, 0)\), with inflection points at \( x = \pm \sqrt{\frac{1}{3}} \). It is concave up on \(\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) and concave down elsewhere. The horizontal asymptote at \( y = 1 \) should be included in the graph.

Key Concepts

Derivatives

Derivatives are a fundamental concept in calculus that help us understand how a function behaves. Essentially, a derivative tells us the rate of change of a function at a given point. For example, the first derivative of a function gives us information on whether a function is increasing or decreasing at that point.

To find the derivative of a function like \( f(x) = \frac{x^2}{1+x^2} \), we apply the quotient rule, because the function is a ratio of two expressions. The quotient rule states that if we have a function \( g(x) = \frac{u(x)}{v(x)} \), then its derivative \( g'(x) \) is given by:

• \( g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2} \)

For \( f(x) \), the first derivative is found to be \( f'(x) = \frac{2x}{(1+x^2)^2} \). This result will help us determine where the function increases or decreases.

Derivatives provide crucial insights, like finding critical points and testing for concavity, which are key topics we'll explore further.

Critical Points

Critical points of a function occur where its first derivative is zero or undefined. They are significant because they may indicate local maxima, minima, or saddle points. Identifying these points helps in understanding the function's overall behavior.

In the case of \( f(x) = \frac{x^2}{1+x^2} \), we set the first derivative \( f'(x) = \frac{2x}{(1+x^2)^2} \) equal to zero to find critical points:

• \( 2x = 0 \) implies \( x = 0 \)

Since the denominator \((1+x^2)^2 \) is never zero, the only critical point is \( x = 0 \). To determine if this critical point is a local maximum or minimum, we examine the sign of the derivative. As the solution explains, \( f'(x) \) is positive for \( x > 0 \) and negative for \( x < 0 \). This means that \( x = 0 \) is a point where the function changes from decreasing to increasing, indicating a local minimum.

Understanding critical points lets us sketch more accurate graphs, showing where the function takes on extreme values.

Concavity

Concavity describes how the slope of a function is changing. A function is concave up if its slope increases, and concave down if its slope decreases. This behavior can be determined by looking at the sign of the second derivative.

For \( f(x) = \frac{x^2}{1+x^2} \), we found the second derivative to be \( f''(x) = \frac{2 - 6x^2}{(1+x^2)^3} \). By setting \( f''(x) = 0 \), we find potential inflection points where concavity changes:

• \( 2 - 6x^2 = 0 \) results in \( x = \pm\sqrt{\frac{1}{3}} \).

Investigating the sign of \( f''(x) \) tells us that the function is concave up on the interval \( \left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \) and concave down outside this interval.

These changes in concavity correspond to the inflection points \( x = \pm\sqrt{\frac{1}{3}} \), where the function transitions between being concave up and concave down. Recognizing these points helps in understanding the full shape of the function's graph.

Horizontal Asymptotes

Horizontal asymptotes indicate where a function approaches a constant value as the independent variable gets very large in either a positive or negative direction. They provide valuable information about the end behavior of a function.

For \( f(x) = \frac{x^2}{1+x^2} \), we calculate the limits as \( x \to \infty \) and \( x \to -\infty \) to determine horizontal asymptotes. The calculations show that:

• \( \lim_{x \to \infty} f(x) = 1 \)
• \( \lim_{x \to -\infty} f(x) = 1 \)

Thus, we find that \( y = 1 \) is a horizontal asymptote of \( f(x) \). This means that as we move far to the right or left along the x-axis, the function's output approaches 1.

Knowing about horizontal asymptotes allows us to predict long-term behavior of the function, which is especially useful when sketching the graph and understanding how the function behaves at extreme values of \( x \).