Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-18, 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

Concave up on \((- fty, -\sqrt[4]{3}) \cup (\sqrt[4]{3}, \ fty)\); concave down on \((-\sqrt[4]{3}, \sqrt[4]{3})\); inflection points at \((\pm \sqrt[4]{3}, \frac{2\sqrt{3}}{3})\).

Step 1: Find the First Derivative

First, we find the first derivative of the function \( f(z) = z^2 - \frac{1}{z^2} \).The derivative of \( z^2 \) is \( 2z \), and the derivative of \( -\frac{1}{z^2} \) is \( 2z^{-3} \). Therefore, the first derivative is:\[ f'(z) = 2z + \frac{2}{z^3} \]

Step 2: Find the Second Derivative

Next, calculate the second derivative to determine concavity. The derivative of \( 2z \) is \( 2 \), and the derivative of \( \frac{2}{z^3} \) is \(-6z^{-4}\). Thus, the second derivative is:\[ f''(z) = 2 - \frac{6}{z^4} \]

Step 3: Determine Sign of the Second Derivative

We determine where \( f''(z) > 0 \) (concave up) and \( f''(z) < 0 \) (concave down). Set \( f''(z) = 2 - \frac{6}{z^4} \) to find when it changes sign:\[ 2 - \frac{6}{z^4} = 0 \]\[ 2 = \frac{6}{z^4} \]\[ z^4 = 3 \]Thus, the critical values are \( z = \pm \sqrt[4]{3} \). Evaluate the sign of \( f''(z) \) around these values.

Step 4: Analyze Intervals for Concavity

Test intervals around \( z = \pm \sqrt[4]{3} \):- For \( z < -\sqrt[4]{3} \): \( f''(z) > 0 \), concave up.- For \( -\sqrt[4]{3} < z < \sqrt[4]{3} \): \( f''(z) < 0 \), concave down.- For \( z > \sqrt[4]{3} \): \( f''(z) > 0 \), concave up.These tests determine the intervals of concavity.

Step 5: Identify Inflection Points

Determine where concavity changes by using the critical points \( z = \pm \sqrt[4]{3} \). Since \( f''(z) \) changes from positive to negative or vice versa at these points, \( z = \pm \sqrt[4]{3} \) are inflection points.Calculate the function values at these points:- At \( z = \sqrt[4]{3} \), \( f(\sqrt[4]{3}) = (\sqrt[4]{3})^2 - \frac{1}{(\sqrt[4]{3})^2} = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \)- Similarly, \( f(-\sqrt[4]{3}) = \frac{2\sqrt{3}}{3} \)Thus, inflection points are \( \left( \sqrt[4]{3}, \frac{2\sqrt{3}}{3} \right) \) and \( \left( -\sqrt[4]{3}, \frac{2\sqrt{3}}{3} \right) \).

Key Concepts

First Derivative

The first derivative of a function gives us valuable information about its slope. It's like taking a snapshot of how steep the function is at any given point. For a function like \( f(z) = z^2 - \frac{1}{z^2} \), the first derivative can be found by applying basic differentiation rules.
Here are the steps:

• The derivative of \( z^2 \) is \( 2z \).
• The derivative of \(-\frac{1}{z^2} \) is \( \frac{2}{z^3} \).

Combining these, the first derivative \( f'(z) \) becomes \( 2z + \frac{2}{z^3} \).
This derivative tells us how the function's values are changing at each point \( z \). It is a crucial first step in analyzing curves, especially when determining where they increase or decrease.

Second Derivative

The second derivative of a function provides insight into the function's concavity, which relates to how the curve is bending. For the function \( f(z) = z^2 - \frac{1}{z^2} \), calculating the second derivative involves differentiating the first derivative again.
Here's how you find it:

• The derivative of \( 2z \) is \( 2 \).
• The derivative of \( \frac{2}{z^3} \) is \(-\frac{6}{z^4} \).

By combining these terms, the second derivative \( f''(z) \) becomes \( 2 - \frac{6}{z^4} \).
This derivative helps determine not just the slope but how curves change direction. Its sign indicates concavity (either concave up or down), playing a pivotal role in identifying inflection points.

Inflection Points

Inflection points are special locations on a curve where the concavity changes. In simpler terms, it's where a curve switches from being 'cup-shaped' to 'cap-shaped' or vice versa.
For our function \( f(z) = z^2 - \frac{1}{z^2} \), we find inflection points by setting the second derivative \( f''(z) = 2 - \frac{6}{z^4} \) to zero and solving for \( z \). This gives us critical points at \( z = \pm \sqrt[4]{3} \).
To confirm they are inflection points, check if \( f''(z) \) changes sign around these values. Indeed, it transitions from positive to negative or negative to positive.
Evaluating the function \( f(z) \) at these points provides their coordinates, \( \left( \sqrt[4]{3}, \frac{2\sqrt{3}}{3} \right) \) and \( \left( -\sqrt[4]{3}, \frac{2\sqrt{3}}{3} \right) \). These are the points of inflection.

Concave Up

A function is concave up where its graph resembles the shape of a cup. This happens when the second derivative is positive.
To find where \( f(z) = z^2 - \frac{1}{z^2} \) is concave up, evaluate \( f''(z) = 2 - \frac{6}{z^4} \) for different intervals.
The intervals around the critical points \( z = \pm \sqrt[4]{3} \) are key:

• For \( z < -\sqrt[4]{3} \), \( f''(z) > 0 \).
• For \( z > \sqrt[4]{3} \), \( f''(z) > 0 \).

In these regions, the graph curves upwards, resembling a bowl or a smile.

Concave Down

When a function's graph looks like an upside-down cup, it is concave down. This occurs where the second derivative is negative.
For \( f(z) = z^2 - \frac{1}{z^2} \), analyze \( f''(z) = 2 - \frac{6}{z^4} \) within intervals around the critical points.
Specifically, for the interval \( -\sqrt[4]{3} < z < \sqrt[4]{3} \), \( f''(z) < 0 \).
This indicates that the graph curves downward, like a frown or an umbrella. Recognizing these intervals provides a better visual understanding of the function's behavior.