Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the Concavity Theorem to determine where the given function is concave up and where it is concave down. Also find all inflection points. $$ f(z)=z^{2}-\frac{1}{z^{2}} $$

Step-by-Step Solution

Answer

The function is concave up for \( |z| > \sqrt[4]{3} \), concave down for \( 0 < |z| < \sqrt[4]{3} \), with inflection points at \( z = \pm \sqrt[4]{3} \).

Step 1: Find the First Derivative

To understand the concavity, first, we need the first derivative of \( f(z) = z^2 - \frac{1}{z^2} \). Using basic differentiation rules, we find:\[ f'(z) = \frac{d}{dz} \left( z^2 \right) - \frac{d}{dz} \left( \frac{1}{z^2} \right) = 2z + \frac{2}{z^3}.\]

Step 2: Find the Second Derivative

Next, calculate the second derivative of \( f(z) \) from \( f'(z) = 2z + \frac{2}{z^3} \):\[ f''(z) = \frac{d}{dz} \left( 2z \right) + \frac{d}{dz} \left( \frac{2}{z^3} \right).\] This results in:\[ f''(z) = 2 - \frac{6}{z^4}.\]

Step 3: Determine Concavity

The concavity of the function is determined by the sign of \( f''(z) \).- If \( f''(z) > 0 \), the function is concave up.- If \( f''(z) < 0 \), the function is concave down.Analyze \( f''(z) = 2 - \frac{6}{z^4} \):- Set \( 2 - \frac{6}{z^4} = 0 \) to find potential points of change in concavity.- Solving gives \( z^4 = 3 \), or \( z = \pm \sqrt[4]{3} \).Now test intervals:- For \( z^4 > 3 \), \( f''(z) > 0 \) (concave up).- For \( z^4 < 3 \), \( f''(z) < 0 \) (concave down).

Step 4: Find Inflection Points

Inflection points occur where \( f''(z) = 0 \) or changes sign.The points \( z = \pm \sqrt[4]{3} \) are potential inflection points, provided \( f(z) \) is defined there.Check each sign change and verify the points result in a change of concavity:- As determined, \( \pm \sqrt[4]{3} \) show a sign change and \( f(z) \) is defined at these points, confirming they are inflection points.

Key Concepts

First Derivative

The first derivative of a function provides essential information about the function's behavior, specifically its slope at any given point. In our exercise, we start with the function \( f(z) = z^2 - \frac{1}{z^2} \). To find the first derivative, we apply the basic rules of differentiation:

• For \( z^2 \), the derivative is \( 2z \).
• For \( -\frac{1}{z^2} \), remember that this is equivalent to \( -z^{-2} \), and its derivative becomes \( 2z^{-3} \).

Combining these, we have the first derivative:\[ f'(z) = 2z + \frac{2}{z^3}. \]This tells us how the function is increasing or decreasing at each value of \( z \). While the first derivative itself is not needed for determining concavity, it provides a foundation for deeper analysis.

Second Derivative

The second derivative of a function is crucial in determining the concavity of a function. For our function \( f(z) \), we already found the first derivative: \( f'(z) = 2z + \frac{2}{z^3} \). Now, we calculate the second derivative, denoted as \( f''(z) \):

• The derivative of \( 2z \) is \( 2 \).
• The derivative of \( \frac{2}{z^3} \), which equates to \( 2z^{-3} \), is \( -6z^{-4} \) or \( -\frac{6}{z^4} \).

Therefore, the second derivative is:\[ f''(z) = 2 - \frac{6}{z^4}. \]This derivative will help us analyze whether parts of the function are concave up or concave down, as well as identify inflection points.

Inflection Points

Inflection points occur where the second derivative equals zero or changes sign, indicating a change in the function's concavity. By setting our second derivative equal to zero:\[ 2 - \frac{6}{z^4} = 0, \]we solve for \( z \) to get:\[ z^4 = 3, \quad \text{or} \quad z = \pm \sqrt[4]{3}. \]These points are candidates for inflection points. Inflection points are valid only if the function is defined at these values and the concavity actually changes from one side to the other. Checking our function, it is defined at these points, and the second derivative changes sign, confirming inflection points at \( z = \pm \sqrt[4]{3} \).

Concave Up

A function is concave up in the regions where its second derivative is positive. Interpret this visually as the graph forming a cup shape. For our function, after finding \( f''(z) = 2 - \frac{6}{z^4} \), it becomes clear that:

• When \( z^4 > 3 \), \( f''(z) > 0 \), implying that the function is concave up.

This generally means that as we check values of \( z \) greater or less than \( \pm \sqrt[4]{3} \), the intervals where this condition is satisfied will exhibit a concave-up shape. Concave-up sections indicate that the slope of the tangent line is increasing.

Concave Down

In contrast to concave up, a function is concave down where its second derivative is negative. This forms an arch shape on the graph. For \( f''(z) = 2 - \frac{6}{z^4} \), the function is concave down when:

• \( z^4 < 3 \), meaning \( f''(z) < 0 \).

These are your intervals leading up to and beyond the inflection points of \( z = \pm \sqrt[4]{3} \). This indicates a decreasing slope of the tangent, thus creating a region where the curve bends downwards. Observing these changes helps determine critical shapes and behaviors of the function's graph.