Problem 18 - 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. $$ G(x)=\arcsin 2 x $$

Step-by-Step Solution

Answer

G(x) is concave up on \( (0, \frac{1}{2}) \), down on \( (\-\frac{1}{2}, 0) \), with inflection point at \( x = 0 \).

Step 1: Check the Domain of the Function

The domain of the function \( G(x) = \arcsin(2x) \) is determined by the range of the arcsin function, which requires \(-1 \leq 2x \leq 1\). Solving for \( x \) gives \(-\frac{1}{2} \leq x \leq \frac{1}{2}\). This is the interval where the function is defined.

Step 2: Find the First Derivative

We need the first derivative of \( G(x) \) to find the concavity. We use the chain rule, where \( u = 2x \), then \( G'(x) = \frac{d}{dx}\arcsin(u) \times \frac{d}{dx}(2x) \). The derivative of \( \arcsin(u) \) is \( \frac{1}{\sqrt{1-u^2}} \). Therefore, \( G'(x) = \frac{2}{\sqrt{1-(2x)^2}} = \frac{2}{\sqrt{1-4x^2}} \).

Step 3: Find the Second Derivative

The second derivative will help determine the concavity. To find \( G''(x) \), we'll differentiate \( G'(x) \). We'll use the quotient rule, where \( u = 2 \) and \( v = \sqrt{1-4x^2} \). d\[/G''(x) = \frac{0 \cdot \sqrt{1-4x^2} - 2 \cdot \left(- \frac{d}{dx}(\sqrt{1-4x^2})\right)}{(\sqrt{1-4x^2})^2} = \frac{2 \cdot (\frac{4x}{\sqrt{1-4x^2}})}{(1-4x^2)} \]. Solve this to get: \( G''(x) = \frac{8x}{(1-4x^2)^{\frac{3}{2}}} \).

Step 4: Determine Intervals of Concavity

Determine the sign of \( G''(x) = \frac{8x}{(1-4x^2)^{\frac{3}{2}}} \). - When \( x > 0 \), \( G''(x) > 0 \), so \( G(x) \) is concave up. - When \( x < 0 \), \( G''(x) < 0 \), so \( G(x) \) is concave down.

Step 5: Find Inflection Points

Inflection points occur where the concavity changes, which is when \( G''(x) = 0 \) or is undefined within the domain. - \( G''(x) = 0 \) when \( 8x = 0 \), which implies \( x = 0 \). So, \( x = 0 \) is an inflection point.

Key Concepts

Concavity Theorem

The Concavity Theorem is a powerful tool used in calculus to describe the curvature or the concavity of a function's graph. This theorem helps us determine where a function is concave up or concave down based on its second derivative.
- A function is said to be **concave up** on an interval if its second derivative, denoted as \( f''(x) \), is greater than zero \( (f''(x) > 0) \). In this case, the function tends to rise upwards.
- Conversely, it is **concave down** if \( f''(x) < 0 \). Here, the graph appears to bend downwards, resembling a frown.
To apply the theorem effectively, we follow specific steps:

• Find the second derivative of the given function.
• Analyze the sign of the second derivative over the domain of the function.
• Identify the intervals of concavity based on the sign of \( f''(x) \).

This approach not only reveals the concavity but also aids in discovering any inflection points where the concavity might change.

Inflection Points

Inflection points are crucial in understanding the change in direction of concavity of a function. An inflection point occurs at a point along the graph of a function where the concavity switches from up to down or vice versa. This happens when the second derivative \( f''(x) \) changes its sign.To locate inflection points, follow these steps:

• First, compute the second derivative of the function.
• Set \( f''(x) = 0 \) and solve the equation to find potential inflection points.
• Check each solution to determine if there’s an actual change in concavity around that point within the function's domain.

It's essential to check that our solutions fall within the domain of the original function, as inflection points outside this domain are irrelevant for practical purposes. When you check these points, ensure there is indeed a transition from concave up to concave down or vice versa.

Derivatives

Derivatives are fundamental in calculus and provide us with a way to understand the behavior of functions in terms of rates of change and slopes. The process of finding a derivative involves differentiation, which tells us how a function's value changes as its input changes.
For the purpose of determining concavity and inflection points:

• The **first derivative** \( f'(x) \) is necessary to determine the slope and the behavior of the function at specific points.
• The **second derivative** \( f''(x) \) goes a step further and helps us understand the concavity—how the function's graph curves.

The first derivative gives us points of tangency, critical points, and potential maximum or minimum values of the function. Calculating the second derivative takes this further by providing insight into how these values are linked with concavity and potential inflection points.
In analyzing the function \( G(x) = \arcsin(2x) \), using the chain rule, we determined its derivatives, leading us to insights about the function’s concavity over its domain.

Domain of Function

Understanding the domain of a function is critical, as it sets the stage for where the function is defined and can be analyzed. For the function \( G(x) = \arcsin(2x) \), it's crucial to establish where the function is valid.
The **domain** is the set of all permissible values of \( x \) for which the function is defined. For trigonometric functions like arcsin:

• The input to the \( \arcsin \) function, here \( 2x \), must satisfy the condition \( -1 \le 2x \le 1 \).
• This can be simplified to \( -\frac{1}{2} \le x \le \frac{1}{2} \), indicating where the function is operational.

Understanding the domain is essential not only for computing derivatives or inflection points accurately, but also for ensuring the results are applicable and meaningful. It specifies the boundaries within which you perform your analysis, ensuring that calculations like concavity and inflection points are valid and accurate within this range.