Problem 37 - Calculus A New Horizon

Question

Give a complete graph of the function, and identify the location of all relative extrema and inflection points. Check your work with a graphing utility. $$\sin ^{2} x, \quad 0 \leq x \leq 2 \pi$$

Step-by-Step Solution

Answer

Relative minima at \( x = 0, \pi, 2\pi \) and maxima at \( x = \frac{\pi}{2}, \frac{3\pi}{2} \). Inflection points at \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 1: Understanding the Function

The given function is \( y = \sin^2(x) \). To find relative extrema and inflection points, we need to find the derivative and analyze where these occur.

Step 2: Find the First Derivative

Use the chain rule to find the derivative of \( \sin^2(x) \). Let \( u = \sin(x) \), then \( y = u^2 \). Thus, \( \frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2\sin(x)\cos(x) = \sin(2x) \).

Step 3: Identify Critical Points for Extrema

Critical points occur where \( \frac{dy}{dx} = \sin(2x) = 0 \). Solve \( \sin(2x) = 0 \) to find critical points: \( 2x = n\pi \), where \( n \) is an integer. Thus, \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \).

Step 4: Use the Second Derivative Test for Extrema

Find the second derivative to apply the second derivative test. The first derivative is \( \sin(2x) \), and its derivative is \( 2\cos(2x) \). Check \( 2\cos(2x) \) at the critical points to determine if they are maxima or minima. If \( 2\cos(2x) > 0 \), it’s a relative minimum; if \( 2\cos(2x) < 0 \), it’s a relative maximum.

Step 5: Analyze and List Extrema

At \( x = 0, \pi, 2\pi \), \( 2\cos(2x) = 2 \), indicating relative minima. At \( x = \frac{\pi}{2}, \frac{3\pi}{2} \), \( 2\cos(2x) = -2 \), indicating relative maxima.

Step 6: Find Inflection Points

Inflection points occur when \( 2\cos(2x) = 0 \), as the second derivative changes sign here. Solve \( \cos(2x) = 0 \) to find inflection points: \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).

Step 7: Graph the Function Using a Graphing Utility

Use a graphing calculator or software to graph \( y = \sin^2(x) \) across the domain \( 0 \leq x \leq 2\pi \). Verify the location of the relative extrema at \( x = 0, \pi/2, \pi, 3\pi/2, 2\pi \) and inflection points at \( x = \pi/4, 3\pi/4, 5\pi/4, 7\pi/4 \).

Key Concepts

Derivative of Trigonometric Functions

When dealing with trigonometric functions like \( y = \sin^2(x) \), understanding how to find the derivative is crucial. Derivatives help us analyze the behavior of the function by determining where it increases or decreases.
To find the derivative of \( \sin^2(x) \), we use the chain rule. This rule allows us to differentiate a composite function by taking the derivative of the outer function and then multiplying it by the derivative of the inner function.
Consider \( u = \sin(x) \). Then \( y = u^2 \). Applying the chain rule, the derivative is: \[ \frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2\sin(x)\cos(x) \].
This can be simplified using the double-angle identity: \( \sin(2x) = 2\sin(x)\cos(x) \). Thus, \( \frac{dy}{dx} = \sin(2x) \) is the first derivative of \( y = \sin^2(x) \). This derivative will help us find the critical points of the function.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points often indicate potential relative maxima or minima for the function.
For the function \( y = \sin^2(x) \), the critical points are found by solving: \( \sin(2x) = 0 \). This happens when \( 2x = n\pi \) for integers \( n \).
Thus, the critical points within the interval \( [0, 2\pi] \) are \( x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \). At these points, the function's slope is zero, which suggests they might be relative extrema. However, to confirm whether these are maxima or minima, we need to perform further testing.

Second Derivative Test

The second derivative test helps determine whether a critical point is a relative maximum or minimum. This test involves analyzing the sign of the second derivative at critical points.
The second derivative of the function \( y = \sin^2(x) \), whose first derivative is \( \sin(2x) \), is \( 2\cos(2x) \).
Here’s how it works:

• If \( 2\cos(2x) > 0 \), the function is concave up, indicating a relative minimum.
• If \( 2\cos(2x) < 0 \), the function is concave down, indicating a relative maximum.

Apply this test to the critical points:
At \( x = 0, \pi, 2\pi \), \( 2\cos(2x) = 2 \), so these points are relative minima.
At \( x = \frac{\pi}{2}, \frac{3\pi}{2} \), \( 2\cos(2x) = -2 \), indicating relative maxima. This analysis helps us identify peaks and troughs in the graph of the function.

Inflection Points

Inflection points occur where the second derivative changes sign. These points indicate a change in the concavity of the function—where the graph changes from concave up to concave down, or vice versa.
For \( y = \sin^2(x) \), the second derivative \( 2\cos(2x) \) equals zero at the inflection points. Solving \( \cos(2x) = 0 \) gives inflection points for \( x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \).
At these points, verify that \( 2\cos(2x) \) indeed changes sign, confirming them as inflection points. These changes in concavity allow us to appreciate the overall shape and character of the graph.
Using these analyses, you can graph and understand the behavior of functions, assisting you with visualizing the peaks, valleys, and flexes of \( y = \sin^2(x) \).