Problem 9 - Calculus for Biology and Medicine

Question

Use the first derivative test and the second derivative test to determine where each function is increasing, decreasing, concave up, and concave down. You do not need to use a graphing calculator for these exercises. $$ y=\frac{1}{x}, x \neq 0 $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, 0)\) and \((0, +\infty)\), concave up on \((0, +\infty)\), and concave down on \((-\infty, 0)\).

Step 1: Find the First Derivative

The function given is \( y = \frac{1}{x} \). First, we find the derivative with respect to \( x \). Using the fact that \( \frac{d}{dx} \frac{1}{x} = -\frac{1}{x^2} \), the first derivative is \( y' = -\frac{1}{x^2} \).

Step 2: Determine Critical Points

Critical points occur where the first derivative is zero or undefined. Since \( y' = -\frac{1}{x^2} \) is never zero but undefined at \( x = 0 \) (where the original function is also undefined), there are no critical points derived from \( y' = 0 \). However, we consider \( x = 0 \) as a point of discontinuity.

Step 3: Apply the First Derivative Test

To determine intervals of increase and decrease, examine the sign of \( y' = -\frac{1}{x^2} \). Note that \( y' < 0 \) for all \( x > 0 \) and \( x < 0 \). Hence, the function is decreasing on both intervals \((-\infty, 0)\) and \((0, +\infty)\).

Step 4: Find the Second Derivative

Compute the second derivative: \( y'' = \frac{d}{dx}(-\frac{1}{x^2}) = \frac{2}{x^3} \).

Step 5: Apply the Second Derivative Test

Examine the sign of the second derivative \( y'' = \frac{2}{x^3} \). For \( x > 0 \), \( y'' > 0 \), indicating the function is concave up. For \( x < 0 \), \( y'' < 0 \), indicating the function is concave down.

Key Concepts

First Derivative Test

The First Derivative Test helps identify where a function is increasing or decreasing. It involves taking the derivative of a function, which shows the slope or rate of change. When the derivative is positive, the function is increasing, and when it is negative, the function is decreasing.

• For the function \( y = \frac{1}{x} \), the first derivative is \( y' = -\frac{1}{x^2} \).
• This derivative is negative for all \( x \) except at points where it is undefined.
• Since \( y' < 0 \) both for \( x > 0 \) and \( x < 0 \), the function is decreasing over the intervals \((-\infty, 0)\) and \((0, +\infty)\).

No solution of \( y' = 0 \) implies there are no standard critical points associated with maxima or minima. However, discontinuity at \( x = 0 \) is crucial for understanding the behavior of the function.

Second Derivative Test

The Second Derivative Test provides information on the concavity of the function, indicating whether it is curving upwards or downwards. This involves taking the second derivative of the function.

• With \( y = \frac{1}{x} \), the second derivative is \( y'' = \frac{2}{x^3} \).
• For \( x > 0 \), \( y'' > 0 \), meaning the function is concave up in this interval.
• For \( x < 0 \), \( y'' < 0 \), indicating the function is concave down in this part of the domain.

Understanding the second derivative helps visualize the curve's bend. Whether the graph of \( y \) looks like a "cup" or a "cap" assists in visualizing potential inflection points or changes in concavity, although they are absent from the intervals provided here, given the continuous negative and positive signs.

Critical Points

Critical points typically occur where the first derivative equals zero or where the derivative is undefined, highlighting potential locations for local maxima, minima, or points of inflection.

• In \( y = \frac{1}{x} \), the first derivative \( y' = -\frac{1}{x^2} \) doesn't provide conventional critical points due to its constant negative slope.
• Usually, critical points help to determine turning points in a function, but since the first derivative here is always negative, the function is exclusively decreasing.
• However, the function is undefined at \( x = 0 \) since both the original and first derivative do not exist here. This discontinuity is essential for understanding the graph's behavior as it approaches this vertical asymptote.

Recognizing these discontinuities is crucial in calculus, revealing essential traits of functions that might not have traditional differential critical points.