Problem 8 - Thomas Calculus

Question

Identify the inflection points and local maxima and minima of the functions graphed in Exercises \(1-8 .\) Identify the intervals on which the functions are concave up and concave down. $$ y=2 \cos x-\sqrt{2} x,-\pi \leq x \leq \frac{3 \pi}{2} $$

Step-by-Step Solution

Answer

Local minima and maxima: \( x = -\frac{3\pi}{4}, -\frac{5\pi}{4} \). Inflection points: \( x = -\frac{\pi}{2}, \frac{\pi}{2} \). Concave up: \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \), concave down: \((-\pi, \frac{\pi}{2})\).

Step 1: Find the first derivative

To determine the local maxima and minima, we first need to find the first derivative of the function. Given the function \( y = 2 \cos x - \sqrt{2} x \), we differentiate it with respect to \( x \) to get:\[ y' = \frac{d}{dx}(2 \cos x - \sqrt{2} x) = -2 \sin x - \sqrt{2}. \]

Step 2: Find critical points using the first derivative

Set the first derivative equal to zero to find the critical points:\[ -2 \sin x - \sqrt{2} = 0. \]Solving for \( \sin x \), we get:\[ \sin x = -\frac{\sqrt{2}}{2}. \]Within the interval \( -\pi \leq x \leq \frac{3\pi}{2} \), this equation holds at \( x = -\frac{3\pi}{4} \) and \( x = -\frac{5\pi}{4} \).

Step 3: Use the second derivative to determine concavity and inflection points

Find the second derivative to analyze concavity and inflection points. From the first derivative \( y' = -2 \sin x - \sqrt{2} \), the second derivative is:\[ y'' = \frac{d}{dx}(-2 \sin x - \sqrt{2}) = -2 \cos x. \]

Step 4: Determine concavity intervals

Analyze the sign of the second derivative to find where the function is concave up or concave down:- \( y'' = -2 \cos x > 0 \) when \( \cos x < 0 \), the function is concave up.- \( y'' = -2 \cos x < 0 \) when \( \cos x > 0 \), the function is concave down.The function is concave up on \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \) and concave down on \( (-\pi, \frac{\pi}{2}) \).

Step 5: Find inflection points

Inflection points occur where the concavity changes, which is at points where the second derivative equals zero:\[ -2 \cos x = 0 \Rightarrow \cos x = 0. \]Within \( -\pi \leq x \leq \frac{3\pi}{2} \), this occurs at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

Key Concepts

Inflection Points

Inflection points are specific points on a curve where the concavity changes. This means the curve changes from bending upwards to downwards (or vice versa). A critical step in identifying inflection points is to first find the second derivative of the function. To locate inflection points, solve for when the second derivative is equal to zero, which indicates potential points of concavity change. However, ensure that the concavity actually changes sign at these points. For the function above, we find that the second derivative is zero at \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \). As a result, these are the inflection points, marking where the concavity changes from concave down to up, or vice versa.

Critical Points

Critical points are where the first derivative of a function is zero or undefined, which are possible locations for local maxima or minima. To find these points, solve \( y' = 0 \). This is because at critical points, the slope of the tangent to the curve is horizontal, indicating a peak or valley on the graph.In the provided exercise, the critical points are found when \(-2 \sin x - \sqrt{2} = 0\), which simplifies to \( \sin x = -\frac{\sqrt{2}}{2} \). Within the given interval, the critical points are at \( x = -\frac{3\pi}{4} \) and \( x = -\frac{5\pi}{4} \). These points are critical for identifying local maxima and minima.

Concavity

Concavity indicates how a function curves. It's determined by the sign of the second derivative of the function. - If the second derivative \( y'' > 0 \), the function is concave up, resembling the shape of a cup or a U.- If \( y'' < 0 \), the function is concave down, like an upside-down cup or an inverted U.For the function \( y = 2\cos x - \sqrt{2}x \), the second derivative is \( y'' = -2\cos x \). This tells us:

• Concave up on intervals where \( \cos x < 0 \), specifically \( \left(\frac{\pi}{2}, \frac{3\pi}{2}\right) \).
• Concave down on intervals where \( \cos x > 0 \), noted as \( (-\pi, \frac{\pi}{2}) \).

Understanding concavity helps recognize the shape and behavior of a graph on different intervals.

Derivatives

Derivatives are fundamental in calculus and represent the rate of change or slope of a function at any given point. The first derivative, \( y' \), provides information about the slope of the tangent line to the function and helps find critical points. The process of differentiation is applied to determine both the first and second derivatives. These derivatives are key to analyzing the function for local peaks, valleys, and inflection points. The first derivative, \( y' = -2 \sin x - \sqrt{2} \), helps locate where peaks or valleys might be by setting it to zero. The second derivative, \( y'' = -2 \cos x \), reveals the curvature direction. Understanding how to differentiate is crucial to solving calculus problems and analyzing functions effectively.