Problem 17 - 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. \(F(x)=2 x^{2}+\cos ^{2} x\)

Step-by-Step Solution

Answer

The function \( F(x) = 2x^2 + \cos^2 x \) is always concave up, and there are no inflection points.

Step 1: Find the First Derivative

To determine where the function is concave up or down, we first find the first derivative of the function, \( F(x) = 2x^2 + \cos^2 x \). We use the chain rule for the cosine squared term: \[ F'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(\cos^2 x) = 4x - 2\cos(x)\sin(x) \ = 4x - \sin(2x).\]

Step 2: Find the Second Derivative

Next, we find the second derivative \( F''(x) \) to determine the concavity. Differentiate \( F'(x) = 4x - \sin(2x) \): \[ F''(x) = \frac{d}{dx}(4x) - \frac{d}{dx}(\sin(2x)) = 4 - 2\cos(2x). \]

Step 3: Determine Concavity Intervals

To find the intervals of concavity, we solve the inequality \( F''(x) = 4 - 2\cos(2x) \) for concavity. When \( F''(x) > 0 \), the function is concave up, and when \( F''(x) < 0 \), it is concave down. Solve:- \( 4 - 2\cos(2x) > 0 \implies \cos(2x) < 2 \). Since \( \cos(2x) \leq 1 \), this is always true for concave up.- \( 4 - 2\cos(2x) < 0 \implies \cos(2x) > 2 \) is impossible because it is never true for \( \cos(2x) \leq 1 \). Hence, no concave down region arises from this inequality.

Step 4: Identify Inflection Points

Inflection points occur where the second derivative changes sign. Given \( F''(x) = 4 - 2\cos(2x) \) and no changes in sign exist besides where it's undefined, test solutions for these undefined crossings which, after analysis for ranges, does not exist at finite numbers. Therefore, there are no real inflection points.

Key Concepts

Inflection Points

Inflection points are fascinating features that help us understand the "shape" of a graph. They are specific points on a curve where the concavity changes. This means the curve switches from being twisted upwards (concave up) to downwards (concave down), or vice versa.
To find these points, we rely on the function's second derivative, which tells us how the slope of the tangent line is changing.

• If the second derivative changes sign at a point, the function is likely crossing an inflection point there.
• It's essential to verify if there are real mathematical values where this switch occurs, as sometimes analysis could lead to a non-existent inflection point.

In the given function, despite solving for potential inflection points, the results show no real inflection points due to the nature of the cosine function, reminding us to carefully evaluate second derivatives for their behavior over real numbers.

Second Derivative Test

The Second Derivative Test is a powerful tool in calculus for analyzing the curvature of a function. Essentially, it helps determine a function's concavity and the nature of its critical points.
The test is simple:

• If the second derivative, \(F''(x)\), is greater than zero at a point, the function is concave up around that point.
• If \(F''(x)\) is less than zero, the function is concave down.
• At the points where the second derivative is zero, you might have inflection points, but further testing is required to confirm this.

In the exercise, we used the Second Derivative Test to identify intervals where the function is concave up. We found that \(4 - 2\cos(2x) > 0\) is always true since \(\cos(2x)\) cannot exceed 1, which automatically nullifies any concave down intervals.

Chain Rule

The Chain Rule is one of the essential rules for differentiation, particularly when dealing with composite functions. It allows us to differentiate a function by recognizing its inner and outer functions.
In the exercise, we applied the chain rule to the \(\cos^2(x)\) component:

• Identify the outer function (squared term) and the inner function (cosine).
• Differentiate the outer function, then multiply it by the derivative of the inner function.

Concisely put, the chain rule guides us in breaking down complex derivatives into manageable parts. As seen with \(\cos^2(x)\), the chain rule is invaluable for finding derivatives whenever you encounter functions nested within each other.

Trigonometric Functions Derivatives

Understanding the derivatives of trigonometric functions enriches our calculus toolbox by ensuring we can work with angles and periodic functions.
Three basic derivatives to remember involve sine, cosine, and their multiples:

• The derivative of \(\sin(x)\) is \(\cos(x)\).
• The derivative of \(\cos(x)\) is \(-\sin(x)\).
• When there's an angle multiple, like \(\sin(2x)\), apply the chain rule to find its derivative \(2\cos(2x)\).

These derivatives were crucial when we differentiated \(\sin(2x)\) in the solution for the exercise. Recognizing these patterns simplifies finding the derivative of trigonometric functions, making calculus that little bit more approachable.