Problem 23 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=\sin ^{2}(x), 0

Step-by-Step Solution

Answer

The function is concave up on \((0, \frac{\pi}{4}) \cup (\frac{3\pi}{4}, \pi)\), concave down on \((\frac{\pi}{4}, \frac{3\pi}{4})\), with inflection points at \(x = \frac{\pi}{4}, \frac{3\pi}{4}\), and a local maximum at \(x = \frac{\pi}{2}\).

Step 1: Find the First Derivative

To begin, we start by finding the first derivative of the function \( f(x) = \sin^2(x) \). We apply the chain rule: \( f'(x) = 2\sin(x) \cdot \cos(x) = \sin(2x) \).

Step 2: Find the Critical Points

To find critical points, we set the first derivative \( f'(x) = 0 \):\[ \sin(2x) = 0 \]This equation holds when \( 2x = n\pi \), where \( n \) is an integer. Given the domain \( 0 < x < \pi \), valid solutions are \( x = \frac{n\pi}{2} \). Thus, the critical points are \( x = \frac{\pi}{2} \).

Step 3: Find the Second Derivative

Next, we find the second derivative of the function: \( f''(x) = 2\cos(2x) \).

Step 4: Determine Concavity and Inflection Points

We check where \( f''(x) \) changes sign to find concavity and inflection points. Solving \( 2\cos(2x) = 0 \) gives \( \cos(2x) = 0 \), i.e., \( 2x = \frac{\pi}{2} + n\pi \). Thus, \( x = \frac{\pi}{4} + \frac{n\pi}{2} \). In the range \( 0 < x < \pi \), this gives inflection points at \( x = \frac{\pi}{4} \text{ and } \frac{3\pi}{4} \). The function is concave up where \( \cos(2x) > 0 \) and concave down where \( \cos(2x) < 0 \). Analyzing intervals, the function is concave up on \( \left(0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right) \) and concave down on \( \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \).

Step 5: Apply the Second Derivative Test to Determine Local Extrema

Utilize the Second Derivative Test at the critical point \( x = \frac{\pi}{2} \):- Calculate \( f''\left(\frac{\pi}{2}\right) = 2\cos(\pi) = -2 \), indicating the function is concave down at this point, and thus, has a local maximum.

Key Concepts

Concavity

Concavity describes how the curve of a function behaves around specific intervals. It helps us know whether a function is bending upwards or downwards within those intervals.
To determine concavity, we often look at the second derivative, denoted as \( f''(x) \). Here's how it works:

• If \( f''(x) > 0 \), the function is concave up. This means the graph looks like a cup and is opening upwards.
• If \( f''(x) < 0 \), the function is concave down. Imagine the graph looking like a frown turning downward.

In our example with \( f(x) = \sin^2(x) \), for \( 0 < x < \pi \), we observe that:

• The function is concave up on \( \left(0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \pi\right) \).
• The function is concave down on \( \left(\frac{\pi}{4}, \frac{3\pi}{4}\right) \).

This understanding helps identify the general shape of the graph across these intervals.

Critical Points

Critical points are key places on a function's graph where the slope is zero or undefined. These points are significant because they can potentially indicate a maximum, minimum, or saddle point. Finding critical points involves the following steps: - Calculate the first derivative, \( f'(x) \). - Set \( f'(x) = 0 \) and solve for \( x \).In our exercise, we found the first derivative to be \( f'(x) = \sin(2x) \). Setting it to zero results in: \[ \sin(2x) = 0 \]Solving this equation within the interval \( 0 < x < \pi \) leads us to critical points at \( x = \frac{\pi}{2} \). At these points, the function's direction can change, making them crucial for further analyses like determining local extrema.

Local Maximum

A local maximum is a point where a function reaches a peak in a small surrounding area. It's like the top of a hill in your neighborhood, higher than all immediately nearby points. To determine if a critical point is a local maximum, the Second Derivative Test is employed:

• Evaluate the second derivative at the critical point, \( f''(x) \).
• If \( f''(x) < 0 \), then there's a local maximum at that point because the graph is concave down at this point.

For our function, \( f(x) = \sin^2(x) \), we calculated \( f''\left(\frac{\pi}{2}\right) = -2 \).
This negative value confirms that the function is indeed concave down at \( x = \frac{\pi}{2} \), resulting in a local maximum there.
This finding is crucial when analyzing function graphs to understand peaks in certain intervals.

Inflection Points

Inflection points are special places on a graph where the concavity changes. They signal where a function transitions from bending up to down, or vice versa. Detecting inflection points involves looking at changes in the sign of the second derivative, \( f''(x) \).Here's how to find them:

• Evaluate \( f''(x) \) and solve \( f''(x) = 0 \) to find potential inflection points.
• Check the intervals around these points to see where \( f''(x) \) changes sign.

For \( f(x) = \sin^2(x) \), with \( f''(x) = 2\cos(2x) \):

• We know \( f''(x) = 0 \) when \( \cos(2x) = 0 \).
• The solutions \( x = \frac{\pi}{4} \) and \( x = \frac{3\pi}{4} \) within \( 0 < x < \pi \) produce inflection points.

These locations mark where the graph's curvature changes, crucial for understanding the entire behavior of the function.