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

Step-by-Step Solution

Answer

The function has no concave up or down intervals, and no inflection points.

Step 1: Find the First Derivative

To apply the Concavity Theorem, start by finding the first derivative of the function. Given \( F(x) = 2x^2 + \cos^2 x \), use the chain rule and power rule to differentiate.\[ F'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}(\cos^2 x) \]\[ F'(x) = 4x - 2\cos(x)\sin(x) \]Simplify using the identity \( 2 \cos(x) \sin(x) = \sin(2x) \), so:\[ F'(x) = 4x - \sin(2x) \]

Step 2: Find the Second Derivative

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

Step 3: Determine Concavity Intervals

To find where the function is concave up or down, solve \( F''(x) > 0 \) and \( F''(x) < 0 \):1. \( F''(x) = 4 - 2\cos(2x) > 0 \) - Rearrange to \( 2\cos(2x) < 4 \), which implies \( \cos(2x) < 2 \). This is always true since \( \cos(2x) \) is bounded between -1 and 1. - So, the function is concave up nowhere under these theoretical conditions.2. \( F''(x) < 0 \) - Rearrange \( 4 - 2\cos(2x) < 0 \) which implies \( \cos(2x) > 2 \) and does not provide feasible solutions under bounded limits of \( \cos(2x) \).Therefore, revisit context: since \( \cos(2x) \) is between -1 and 1, check directly at roots. No real intervals strictly meet \( < 0 \).

Step 4: Find Inflection Points

Inflection points occur where the concavity changes sign, or \( F''(x) = 0 \). Solve:\[ 4 - 2\cos(2x) = 0 \]\[ 2\cos(2x) = 4 \]\[ \cos(2x) = 2 \]This condition never occurs because \( \cos(2x) \) always lies between -1 and 1, meaning there are no inflection points for this function.

Key Concepts

First Derivative

In calculus, the first derivative of a function, often denoted as \( F'(x) \), reveals information about the function's rate of change. It is essential when applying the Concavity Theorem as the first step involves differentiating the original function. For the exercise function, we start with \( F(x) = 2x^2 + \cos^2 x \). To find \( F'(x) \), the chain and power rules are applied:

• The derivative of \( 2x^2 \) using the power rule is \( 4x \).
• Using the chain rule for \( \cos^2 x \), the derivative is \(-2\cos(x)\sin(x)\), which simplifies using the identity \( 2\cos(x)\sin(x) = \sin(2x) \) to yield \(-\sin(2x) \).

Thus, the first derivative is: \[ F'(x) = 4x - \sin(2x) \] This expression will help us in understanding how the function's values increase or decrease.

Second Derivative

The second derivative, denoted as \( F''(x) \), informs us about the curvature or concavity of the function. It is crucial to check changes in concavity and find inflection points. To obtain \( F''(x) \), we differentiate \( F'(x) = 4x - \sin(2x) \).

• The derivative of \( 4x \) is straightforwardly \( 4 \).
• The derivative \( \frac{d}{dx}(\sin(2x)) \) involves the chain rule again, yielding \( 2\cos(2x) \).

Thus, the second derivative is: \[ F''(x) = 4 - 2\cos(2x) \] This derivative is vital for finding intervals where the function is concave up or down, as well as any potential inflection points.

Inflection Points

Inflection points are where a function transitions between concave up and concave down. These occur where the second derivative equals zero, signifying a change in curvature. For our function, we set:\[ F''(x) = 4 - 2\cos(2x) = 0 \] Solving, we find:\[ 2\cos(2x) = 4 \] \[ \cos(2x) = 2 \] This result, however, is not possible since the cosine function's range is limited to values between -1 and 1. Therefore, the equation has no solutions in the real-number domain, indicating that there are no inflection points for this function. This aspect emphasizes that not all functions have inflection points, especially if their mathematical conditions fall out of real bounds.

Concavity Intervals

Understanding concavity intervals involves determining where a function is concave up or concave down based on the sign of its second derivative. Concave up implies \( F''(x) > 0 \), and concave down implies \( F''(x) < 0 \). For this specific function:

• For \( F''(x) > 0 \): Solving \[ 4 - 2\cos(2x) > 0 \] gives \[ \cos(2x) < 2 \]. Since the maximum value for \( \cos \) is 1, this inequality is always true, but it doesn’t lead to any real interval solutions due to identical conditions.
• For \( F''(x) < 0 \): Solving \[ 4 - 2\cos(2x) < 0 \] leads to \[ \cos(2x) > 2 \], which is also not possible for the \( \cos \) range.

Thus, the function's interval exploration under theoretical bounds provides no traditional concave up or down segments, reflecting the nuanced nature of cosines in the function formula. It's integral to verify mathematical confines when assessing concavity.