Problem 232 - Calculus Volume 1

Question

For the following exercises, determine $$ \begin{array}{l}{\text { a. intervals where } f \text { is increasing or decreasing, }} \\ {\text { b. local minima and maxima of } f} \\ {\text { c. intervals where } f \text { is concave up and concave }} \\ {\text { down, and }} \\ {\text { d. the inflection points of } f \text { . Sketch the curve, then }} \\ {\text { use a calculator to compare your answer. If you }} \\\ {\text { cannot determine the exact answer analytically, use }} \\ {\text { a calculator. }}\end{array} $$ $$ f(x)=x+\sin (2 x) \text { over } x=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] $$

Step-by-Step Solution

Answer

Increasing: \((-\frac{\pi}{3}, \frac{\pi}{3})\), decreasing elsewhere. Local max: \(x = \frac{\pi}{3}\), min: \(x = -\frac{\pi}{3}\). Concave up: \([-\frac{\pi}{2}, 0]\), down: \((0, \frac{\pi}{2}]\). Inflection at \(x = 0\).

Step 1: Find the derivative

To determine where the function is increasing or decreasing, we first calculate the derivative of the function. Given the function \( f(x) = x + \sin(2x) \), the derivative is found using basic differentiation rules:\[ f'(x) = 1 + 2\cos(2x) \]

Step 2: Determine critical points

Critical points occur where the derivative is zero or undefined. Set the derivative equal to zero and solve:\[ 1 + 2\cos(2x) = 0 \]\[ 2\cos(2x) = -1 \]\[ \cos(2x) = -\frac{1}{2} \]The solutions in the interval \([-\frac{\pi}{2}, \frac{\pi}{2}]\) for \(2x\) are \(\frac{2\pi}{3}\) and \(-\frac{2\pi}{3}\). Solve for \(x\):\[ x = \frac{\pi}{3} \text{ and } x = -\frac{\pi}{3} \]

Step 3: Test intervals for increasing/decreasing

Evaluate \( f'(x) \) around each critical point to determine if the function is increasing or decreasing. - For \(x\) in \([-\frac{\pi}{2}, -\frac{\pi}{3})\), pick \(x = -\frac{\pi}{2}\): \( f'(-\frac{\pi}{2}) = 1 + 2\cos(-\pi) = -1 \), which is negative (decreasing).- For \(x\) in \((-\frac{\pi}{3}, \frac{\pi}{3})\), pick \(x = 0\): \( f'(0) = 1 + 2 = 3 \), which is positive (increasing).- For \(x\) in \((\frac{\pi}{3}, \frac{\pi}{2}]\), pick \(x = \frac{\pi}{2}\): \( f'(\frac{\pi}{2}) = 1 - 2 = -1 \), which is negative (decreasing).

Step 4: Determine local maxima and minima

Based on the sign changes in \( f'(x) \):- \( f(x) \) has a local minimum at \(x = -\frac{\pi}{3}\) because \( f'(x) \) changes from negative to positive.- \( f(x) \) has a local maximum at \(x = \frac{\pi}{3}\) because \( f'(x) \) changes from positive to negative.

Step 5: Find the second derivative for concavity

Calculate the second derivative to determine concavity:\[ f''(x) = -4\sin(2x) \]

Step 6: Test intervals for concavity

Determine where \( f''(x) \) is greater than or less than zero:- \( f''(x) > 0 \) (concave up) when \(-4\sin(2x) > 0 \). This occurs when \( \sin(2x) < 0 \) or, equivalently, in the intervals \([ -\frac{\pi}{2}, 0] \).- \( f''(x) < 0 \) (concave down) when \(-4\sin(2x) < 0 \). This occurs when \( \sin(2x) > 0 \), or in the interval \((0, \frac{\pi}{2}] \).

Step 7: Determine inflection points

Inflection points occur where \( f''(x) \) changes sign. These will be the solutions to \( \sin(2x) = 0 \), which occur at \(x = 0\).

Key Concepts

Derivatives

Derivatives play a crucial role in understanding the behavior of a function. In calculus, a derivative represents the rate at which a function is changing at any given point. For functions like \( f(x) = x + \sin(2x) \), we use derivatives to analyze how the function increases or decreases.
By taking the derivative of \( f(x) \), which is \( f'(x) = 1 + 2\cos(2x) \), we can determine points where the slope of the function is zero or undefined. This helps identify critical changes in the function's behavior.
Calculating and interpreting the derivative is foundational in predicting how functions behave, especially in intervals where they increase or decrease. This is why derivatives are such a fundamental concept in calculus.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. They signify potential maximums, minimums, or inflection points. For the function \( f(x) = x + \sin(2x) \), the critical points are found by setting \( f'(x) = 1 + 2\cos(2x) = 0 \).
Solving this equation, we find the critical points at \( x = \frac{\pi}{3} \) and \( x = -\frac{\pi}{3} \).
These points are essential for analyzing a function because they help determine where the function changes direction from increasing to decreasing or vice versa. Checking around these critical points also allows us to classify them as local maxima or minima, which are important in sketching graphs and understanding function behavior.

Concavity

The concept of concavity refers to whether a function curves upwards or downwards. It’s essential for understanding the shape and feel of a graph. To examine concavity, we look at the second derivative of a function.
For the function \( f(x) = x + \sin(2x) \), the second derivative is \( f''(x) = -4\sin(2x) \).
A positive second derivative indicates that the function is concave up (shaped like a cup), while a negative second derivative indicates that the function is concave down (shaped like a cap).
Understanding concavity helps us predict not only how a graph looks but also where important transitions or bending points, called inflection points, occur in the function.

Inflection Points

Inflection points are where a function changes concavity – from concave up to concave down or vice versa. They are crucial in altering the graph's curvature.
For the function \( f(x) = x + \sin(2x) \), inflection points occur where the second derivative \( f''(x) = -4\sin(2x) \) changes sign.
By solving \( \sin(2x) = 0 \), we find that the inflection point is at \( x = 0 \).
Identifying inflection points is important for understanding the overall structure of a graph and for accurately sketching how it bends or twists between intervals. This insight is valuable for more advanced analyses, such as predicting behavior in real-world applications modeled by such functions.