Problem 35 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\frac{x}{x-1}, \quad x \neq 1 $$ (a) Show that $$ \lim _{x \rightarrow-\infty} f(x)=1 $$ and $$ \lim _{x \rightarrow+\infty} f(x)=1 $$ That is, show that \(y=1\) is a horizontal asymptote of the curve \(y=\frac{x}{x-1}\) (b) Show that $$ \lim _{x \rightarrow 1^{-}} f(x)=-\infty $$ and $$ \lim _{x \rightarrow 1^{+}} f(x)=+\infty $$ That is, show that \(x=1\) is a vertical asymptote of the curve \(y=\frac{x}{x-1}\) (c) Determine where \(f(x)\) is increasing and where \(i t\) is decreasing. Does \(f(x)\) have local extrema? (d) Determine where \(f(x)\) is concave up and where it is concave down. Does \(f(x)\) have inflection points? (e) Sketch the graph of \(f(x)\) together with its asymptotes.

Step-by-Step Solution

Answer

(a) \( y=1 \) is a horizontal asymptote. (b) \( x=1 \) is a vertical asymptote. (c) Increasing everywhere. No local extrema. (d) Concave up on \((-\infty, 1)\), down on \((1, \infty)\). No inflection points.

Step 1: Find Horizontal Asymptotes

To find horizontal asymptotes, we analyze \( \lim_{x \to -\infty} f(x) \) and \( \lim_{x \to +\infty} f(x):\)The function is \( f(x) = \frac{x}{x-1} \).For \( x \to +\infty \) or \( x \to -\infty \), the dominant term in both the numerator and the denominator is \( x \), thus:\[\lim_{x \to \pm\infty} \frac{x}{x-1} = \lim_{x \to \pm\infty} \frac{x}{x(1 - \frac{1}{x})} = \lim_{x \to \pm\infty} \frac{1}{1 - \frac{1}{x}} = 1 \]Therefore, \( y = 1 \) is a horizontal asymptote.

Step 2: Find Vertical Asymptote

Vertical asymptotes occur where the denominator equals zero, provided the numerator is non-zero.For \( f(x) = \frac{x}{x-1} \), the denominator is zero when \( x - 1 = 0 \), so \( x = 1 \).Evaluate the limits near \( x = 1 \):- \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{x}{x-1} = -\infty \) as \( x \) approaches 1 from the left.- \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x}{x-1} = +\infty \) as \( x \) approaches 1 from the right.Thus, \( x = 1 \) is a vertical asymptote.

Step 3: Determine Increasing/Decreasing Intervals

Find the derivative to determine where \( f(x) \) is increasing/decreasing:\[ f'(x) = \frac{d}{dx}\left(\frac{x}{x-1}\right) = \frac{1}{(x-1)^2} \]Since \( f'(x) > 0 \) for all \( x eq 1 \), the function is increasing on all intervals where it is defined.

Step 4: Identify Local Extrema

Since \( f'(x) > 0 \) everywhere in its domain, \( f(x) \) is strictly increasing and thus has no local maxima or minima.

Step 5: Determine Concavity

Find the second derivative:\[ f''(x) = \frac{d}{dx}\left(\frac{1}{(x-1)^2}\right) = \frac{-2}{(x-1)^3} \]- \( f''(x) < 0 \) when \( x > 1 \), so \( f(x) \) is concave down there.- \( f''(x) > 0 \) when \( x < 1 \), so \( f(x) \) is concave up there.

Step 6: Identify Inflection Points

The inflection point occurs where the concavity changes, but there is no \( x \) value where this can occur as \( x = 1 \) is not in the domain. Therefore, no inflection points exist.

Step 7: Sketch the Graph

To sketch the graph:1. Use the asymptotes: \( y = 1 \) (horizontal), \( x = 1 \) (vertical).2. Note \( f(x) \) is strictly increasing.3. Show concavity changes on either side of \( x = 1 \): concave up for \( x < 1 \) and concave down for \( x > 1 \).4. The graph approaches the line \( y = 1 \) as \( x \to \pm \infty \) and diverges along \( x = 1 \).

Step 8: Concluding the Analysis

The function \( f(x) = \frac{x}{x-1} \) has a horizontal asymptote at \( y = 1 \) and a vertical asymptote at \( x = 1 \). It is strictly increasing across its domain with no local extrema. The function is concave up on \( (-\infty, 1) \) and concave down on \( (1, \infty) \) with no inflection points.

Key Concepts

Increasing and Decreasing Functions

In calculus, understanding where a function is increasing or decreasing is key to analyzing its behavior. For the function \( f(x) = \frac{x}{x-1} \), we determine this by finding the first derivative, \( f'(x) \). Calculating the derivative gives us:
\[ f'(x) = \frac{1}{(x-1)^2} \]
This result tells us that \( f'(x) \) is positive for all \( x eq 1 \). Whenever a derivative is positive, the function is increasing. Therefore, \( f(x) \) is strictly increasing in its whole domain, excluding the point \( x = 1 \) where it is undefined.

• The function does not have any local maximum or minimum since it doesn't "turn around" at any point in its domain.
• This consistent increase signifies each subsequent value of \( f(x) \) is always larger than the preceding one as \( x \) moves away from \( x = 1 \).

Concavity and Inflection Points

To determine concavity, we explore the second derivative of the function. For \( f(x) = \frac{x}{x-1} \), the second derivative is:
\[ f''(x) = \frac{-2}{(x-1)^3} \]
Concavity tells us how the function curves. If \( f''(x) > 0 \), the function is concave up, resembling a cup shape. Conversely, if \( f''(x) < 0 \), it indicates concave down, similar to an upside-down cup.

• In this case, for \( x < 1 \), \( f''(x) > 0 \), meaning \( f(x) \) is concave up.
• For \( x > 1 \), \( f''(x) < 0 \), indicating that \( f(x) \) is concave down.

Inflection points occur where the concavity changes. However, \( x = 1 \) is not in the domain of \( f(x) \), so, unfortunately, there are no inflection points for this function. This absence of inflection points implies the concavity transitions occur outside the realm of the defined function.

Calculus Graph Sketching

Graph sketching is the art of combining all the information about a function to visualize its behavior comprehensively. For \( f(x) = \frac{x}{x-1} \), we consider:

• The horizontal asymptote given by \( y = 1 \). This is found as \( x \to \pm \infty \), where the function approaches this line without touching it.
• The vertical asymptote at \( x = 1 \), where the function diverges either to \( +\infty \) or \(-\infty \) depending on the direction of approach.
• With the fact that \( f(x) \) is strictly increasing, it ensures a smooth transition from left to right over each section of its domain.
• The concavity, which is up when \( x < 1 \) and down when \( x > 1 \), gives further detail to the graph's curvature.

When sketching, draw a curve that approaches the horizontal asymptote, diverges at the vertical asymptote, and smoothly transitions based on increasing behavior and concavity. This helps learners visualize how calculus concepts translate into real-world representations on a graph.