Problem 17 - Calculus for Biology and Medicine

Question

Let $$f(x)=\frac{1}{x(x+1)}, \quad x \neq 0,-1$$ (a) Show that \(x=0\) and \(x=-1\) are vertical asymptotes. (b) Determine where \(f(x)\) is increasing and where it is decreasing. Does \(f(x)\) have local extrema? (c) Determine where \(f(x)\) is concave up and where it is concave down. Does \(f(x)\) have inflection points? (d) What is the behavior of the function as \(x \rightarrow \pm \infty\) ? (e) Sketch the graph of \(f(x)\) together with its asymptotes.

Step-by-Step Solution

Answer

(a) Asymptotes at \(x=0, x=-1\); (b) Increase on \((-\infty,-1/2)\), decrease on \((-1/2,\infty)\); max at \(x=-1/2\); (c) Concave up on \((-1/3,\infty)\), down on \((-\infty,-1/3)\); inflection at \(x=-1/3\); (d) As \(x\rightarrow \infty\), \(f(x)\rightarrow 0\).

Step 1: Vertical Asymptotes

The function \( f(x) = \frac{1}{x(x+1)} \) has vertical asymptotes where the denominator is zero. Thus, we solve \( x(x+1) = 0 \).This implies:- \( x = 0 \)- \( x = -1 \)These values make the denominator zero, thus \( x = 0 \) and \( x = -1 \) are vertical asymptotes.

Step 2: Determine Intervals of Increase/Decrease

To find where \( f(x) \) is increasing or decreasing, we first find the derivative \( f'(x) \). Using the quotient rule, we get:\[ f'(x) = \frac{-(2x+1)}{x^2(x+1)^2} \]- \( f'(x) > 0 \) when \(-2x-1 < 0\), which simplifies to \( x < -\frac{1}{2} \)- \( f'(x) < 0 \) when \( x > -\frac{1}{2} \)Thus, \( f(x) \) is increasing on \( (-\infty, -\frac{1}{2}) \) and decreasing on \( (-\frac{1}{2}, \infty) \).

Step 3: Check for Local Extrema

A local extremum occurs where \( f'(x) = 0 \). Solving \( -2x-1 = 0 \) gives \( x = -\frac{1}{2} \). This is a maximum since \( f'(x) \) changes from positive to negative at this point.

Step 4: Concavity and Inflection Points

To determine concavity, find the second derivative \( f''(x) \). Calculating, we have:\[ f''(x) = \frac{2(3x+1)}{x^3(x+1)^3} \]- \( f''(x) > 0 \) when \( 3x+1 > 0 \), or \( x > -\frac{1}{3} \) (concave up)- \( f''(x) < 0 \) when \( x < -\frac{1}{3} \) (concave down)An inflection point occurs where \( f''(x) = 0 \) at \( x = -\frac{1}{3} \).

Step 5: Behavior at Infinity

As \( x \to \pm \infty \), the term \( \frac{1}{x(x+1)} \to 0 \). Thus \( f(x) \) tends to zero, indicating horizontal asymptote at \( y = 0 \).

Step 6: Graph Sketch

Using the information:- Vertical asymptotes at \( x = 0 \) and \( x = -1 \)- Horizontal asymptote at \( y = 0 \)- Increasing on \( (-\infty, -\frac{1}{2}) \), decreasing on \( (-\frac{1}{2}, \infty) \)- Inflection point at \( x = -\frac{1}{3} \)The graph should reflect these characteristics, bending around the asymptotes and showing the increasing/decreasing behavior.

Key Concepts

Vertical Asymptotes

Understanding vertical asymptotes is key when analyzing rational and certain types of transcendental functions in calculus. They occur at points where the function is undefined, thus indicating a value that approaches infinity. For the function given as \( f(x) = \frac{1}{x(x+1)} \), we find the vertical asymptotes by identifying where the denominator equals zero. This entails solving the equation \(x(x+1) = 0\), which gives us two solutions: \( x = 0 \) and \( x = -1 \). Since the denominator becomes zero at these points, they are locations of vertical asymptotes.

The impact of a vertical asymptote is significant, as it denotes a boundary the graph of the function cannot cross. Near these points, the function's value will skyrocket towards positive or negative infinity. Such behavior helps when sketching graphs or discussing the limits and continuity of functions in calculus for biology.

Concavity and Inflection Points

Concavity refers to the curvature of a function's graph – whether it curves upwards or downwards. To determine concavity, we calculate the second derivative \( f''(x) \). For our function \( f(x) = \frac{1}{x(x+1)} \), the second derivative is \( f''(x) = \frac{2(3x+1)}{x^3(x+1)^3} \). The sign of \( f''(x) \) reveals the nature of concavity:

• \( f''(x) > 0 \): The graph is concave up, resembling the shape of a smiley face. This is true when \( 3x+1 > 0 \), or \( x > -\frac{1}{3} \).
• \( f''(x) < 0 \): The graph is concave down, resembling a frown. Occurring when \( x < -\frac{1}{3} \).

an inflection point occurs where the concavity changes sign, indicating a point where the curvature shifts from concave up to concave down, or vice versa. At such points, the second derivative equals zero. For \( f(x) \), the inflection point is found at \( x = -\frac{1}{3} \). Recognizing concavity and inflection points allows us to understand how the biological data modeled by these functions transitions and changes, providing vital insights for interpretation.

Graph Sketching

Graph sketching synthesizes all components of calculus analysis — vertical asymptotes, concavity, increasing or decreasing behavior, and behavior at infinity — to provide a visual representation. When sketching \( f(x) = \frac{1}{x(x+1)} \), start by plotting crucial elements:

• Vertical asymptotes at \( x = 0 \) and \( x = -1 \).
• A horizontal asymptote at \( y = 0 \) as \( x \to \pm \infty \).
• Increase observed on \( (-\infty, -\frac{1}{2}) \) and decrease on \( (-\frac{1}{2}, \infty) \) due to sign changes in the first derivative.
• Inflection point at \( x = -\frac{1}{3} \), marking where the graph shifts concavity.

By linking these aspects, you can draw a more accurate graph showing distinctive growth features and shifts in curve, crucial for biological studies that may rely on such visual analysis to infer trends and project future patterns.