Problem 25 - Calculus of a Single Variable

Question

Finding Points of Inflection In Exercises \(15-30\) , find the points of inflection and discuss the concavity of the graph of the function. $$ f(x)=\sin \frac{x}{2}, \quad[0,4 \pi] $$

Step-by-Step Solution

Answer

The points of inflection of the function \(f(x)=\sin \frac{x}{2}\) over the interval \([0,4 \pi]\) are at \(x = 0\) and \(x = 4\pi\). On the interval \([0,2 \pi]\), \(f(x)\) is concave up. The function \(f(x)\) is concave down on the interval \((2 \pi, 4\pi]\).

Step 1: Find the Derivative of \(f(x)\)

The first derivative of \(f(x) = \sin \frac{x}{2}\) is found using the chain rule. So, \(f'(x) = \frac{1}{2} \cos \frac{x}{2}\).

Step 2: Find the Second Derivative of \(f(x)\)

The second derivative of \(f(x)\) is just the derivative of \(f'(x)\). So, \(f''(x) = -\frac{1}{4} \sin \frac{x}{2}\).

Step 3: Find Potential Points of Inflection

The potential points of inflection occur where \(f''(x) = 0\). Solve \(-\frac{1}{4} \sin \frac{x}{2} = 0\), which yields \(x = 4n\pi\) where \(n\) is an integer. We only need the values within our given domain [0,4π], which are 0 and 4π.

Step 4: Discuss Concavity

For all x where \(x > 8n\pi\) and \(x < (8n+4)\pi\), \(\sin \frac{x}{2}\) is positive, so \(f(x)\) is concave up. For all x where \(x > (8n+4)\pi\) and \(x < (8n+8)\pi\), \(\sin \frac{x}{2}\) is negative, so \(f(x)\) is concave down.

Key Concepts

Concavity

Understanding concavity is key to analyzing the shape and behavior of a graph. A function is concave up when its graph holds the shape of a cup, and this happens when the second derivative of the function is positive. Conversely, a function is concave down when the graph shapes like a frown, indicating that the second derivative is negative.

In our original problem, the concavity of the function is determined by the sign of the second derivative, denoted as \( f''(x) \). By examining \( f''(x) = -\frac{1}{4} \sin \frac{x}{2} \), we determine where the function changes concavity.

• If \( \sin \frac{x}{2} > 0 \), the function is concave down.
• If \( \sin \frac{x}{2} < 0 \), the function is concave up.

By looking at the intervals where the trigonometric function \( \sin \frac{x}{2} \) switches signs within \([0, 4\pi] \), we identify regions of concavity shifts and potential points of inflection.

Chain Rule

The chain rule is a fundamental tool in calculus used for finding the derivative of composite functions. When you have a function that is nested inside another, the chain rule helps by focusing on how the change in the inner function affects the outer one.
To apply the chain rule, you multiply the derivative of the outer function by the derivative of the inner function. In our exercise, the function \( f(x) = \sin \frac{x}{2} \) is a composition of the outer function \( \sin(u) \) and the inner function \( u = \frac{x}{2} \).

When applying the chain rule, we first derive the sine function, which gives us \( \cos(u) \), and then multiply by the derivative of \( u \), which is \( \frac{1}{2} \). Thus, the derivative is \( f'(x) = \frac{1}{2} \cos \frac{x}{2} \). This step is crucial for determining the first derivative and allows us to move forward to find the second derivative.

Second Derivative

The second derivative of a function provides insight into the function's concavity and potential inflection points. It is the derivative of the first derivative, further capturing how the rate of change itself is changing. A positive second derivative suggests that the function is curving upwards (concave up), while a negative one indicates downward curvature (concave down).

For the function \( f(x) = \sin \frac{x}{2} \), we found the first derivative to be \( f'(x) = \frac{1}{2} \cos \frac{x}{2} \). Differentiating this once more, we arrive at the second derivative, \( f''(x) = -\frac{1}{4} \sin \frac{x}{2} \). This second derivative allows us to understand shifts in concavity and identify inflection points where \( f''(x) \) changes its sign, which are key for fully analyzing the graph's behavior.

Trigonometric Functions

Trigonometric functions are pivotal in understanding periodic behavior in functions, particularly in the study of waves and oscillatory motions. Sine, cosine, and tangent are the primary functions, each with their own unique properties.

In this specific exercise, \( \sin \frac{x}{2} \) is interesting because of its periodic nature, which affects how concavity changes over the specified interval. As \( x \) progresses from \( 0 \) to \( 4\pi \), the sine function completes a few cycles, making it essential to understand its periodicity.

• \( \sin \) is periodic with a fundamental period of \( 2\pi \).
• The argument \( \frac{x}{2} \) modifies this period, lengthening it to \( 4\pi \) for \( \sin \frac{x}{2} \).

Recognizing these trigonometric properties helps in accurately predicting where our function changes concavity, and thus identifying inflection points becomes more intuitive.