Problem 15 - 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 it 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

The horizontal asymptote is at \(y = 1\), vertical asymptote at \(x = 1\), the function decreases everywhere and has no extrema or inflection points.

Step 1: Determine Horizontal Asymptotes

To find the horizontal asymptote, evaluate the limits at infinity. For \( \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} \frac{x}{x-1} \), factor out \(x\) in the denominator to get:\[ \lim_{x \to -\infty} \frac{x}{x \left(1 - \frac{1}{x}\right)} \approx \lim_{x \to -\infty} \frac{1}{1 - \frac{1}{x}} = 1 \]Similarly, for \( \lim_{x \to \infty} f(x) \):\[ \lim_{x \to \infty} \frac{x}{x-1} \approx \lim_{x \to \infty} \frac{1}{1 - \frac{1}{x}} = 1 \]Since both limits approach 1, the horizontal asymptote is \(y = 1\).

Step 2: Determine Vertical Asymptotes

Evaluate the limits as \(x\) approaches 1 from the left and right.For \( \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} \frac{x}{x-1} \), as \(x\) approaches 1 from the left, \(x-1\) becomes a small negative number, making the fraction approach \(-\infty\).For \( \lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} \frac{x}{x-1} \), as \(x\) approaches 1 from the right, \(x-1\) becomes a small positive number, making the fraction approach \(+\infty\). Thus, there is a vertical asymptote at \(x = 1\).

Step 3: Find Intervals of Increase and Decrease

Find the derivative \(f'(x)\) to determine where \(f(x)\) is increasing or decreasing:\[ f(x) = \frac{x}{x-1} \implies f'(x) = \frac{-1}{(x-1)^2} \]Since \(f'(x) < 0\) for all \(x eq 1\), the function is decreasing on \((-\infty, 1) \cup (1, \infty)\). Because \(f'(x)\) never equals zero or changes sign, there are no local extrema.

Step 4: Determine Concavity and Inflection Points

Find the second derivative \(f''(x)\):\[ f''(x) = \frac{2}{(x-1)^3} \]Since \(f''(x) > 0\) for \(x > 1\), \(f(x)\) is concave up on \((1, \infty)\).Since \(f''(x) < 0\) for \(x < 1\), \(f(x)\) is concave down on \((-\infty, 1)\). There is no point where \(f''(x) = 0\), so there are no inflection points.

Step 5: Graph the Function and Identify Asymptotes

Using the information from the previous steps, sketch the graph:- The horizontal asymptote is \(y = 1\).- The vertical asymptote is \(x = 1\).- The curve decreases on \((-\infty, 1)\) and \((1, \infty)\).- It is concave down on \((-\infty, 1)\) and concave up on \((1, \infty)\).- No local extrema or inflection points occur. Plot these characteristics accordingly to visualize \(f(x)\).

Key Concepts

Horizontal Asymptote

A horizontal asymptote is a horizontal line that a function approaches as the input grows larger or smaller indefinitely. For the function \(f(x) = \frac{x}{x-1}\), it's important to determine its behavior as \(x\) approaches positive or negative infinity. In this case, both \( \lim_{x \to -\infty} f(x) \) and \( \lim_{x \to \infty} f(x) \) are equal to 1. This means that no matter how large or small \(x\) gets, the value of \(f(x)\) approaches 1. Thus, the horizontal asymptote of the function is the line \(y = 1\).
This tells us that the graph of the function gets closer and closer to this line as we move left or right, but it never actually touches the line.

Vertical Asymptote

Vertical asymptotes occur where the function grows without bound as \(x\) approaches a particular value. For \(f(x) = \frac{x}{x-1}\), as \(x\) approaches 1 from the left, the function tends towards \(-\infty\), and as \(x\) approaches 1 from the right, the function tends towards \(+\infty\).
This dramatic behavior happens because the denominator becomes very small, leading to an infinitely large absolute value of \(f(x)\). Therefore, the line \(x = 1\) is a vertical asymptote. This tells us that the graph of the function acts like a boundary wall at \(x = 1\), where it cannot cross but instead goes off towards infinity in the positive and negative directions as it gets closer to this wall.

Concavity

Concavity describes how the slope of a function behaves. It tells us how the graph bends or curves. To determine the concavity of \(f(x) = \frac{x}{x-1}\), we find the second derivative \(f''(x)\). In this case, \(f''(x) = \frac{2}{(x-1)^3}\).

• For \(x > 1\), \(f''(x) > 0\), meaning the graph is concave up. The curve bends upwards, forming a cup shape in this interval.
• For \(x < 1\), \(f''(x) < 0\), indicating the graph is concave down. Here, the curve bends downwards like an upside-down cup.

Concavity does not change as there are no inflection points where the concavity shifts direction, which means no \(x\) values make \(f''(x)\) equal zero.

Increase and Decrease

To know whether \(f(x)\) is increasing or decreasing, we look at the first derivative \(f'(x)\). For \(f(x) = \frac{x}{x-1}\), the derivative is \(f'(x) = \frac{-1}{(x-1)^2}\).

• Since \(f'(x) < 0\) everywhere except \(x = 1\), the function is always decreasing on the intervals \((-\infty, 1)\) and \((1, \infty)\).

There are no intervals where \(f(x)\) is increasing, meaning \(f(x)\) does not have any local maximum or minimum points. In simpler terms, the graph of \(f(x)\) is perpetually going down when moving from left to right, except it jumps from \(-\infty\) to \(+\infty\) at the vertical asymptote \(x=1\).