Chapter 5: Problem 19
Find the relative extrema using both the first and second derivative tests.
$$f(x)=\sin ^{2} x, \quad 0
Short Answer
Expert verified
Local maxima at \(x = \frac{\pi}{2}, \frac{3\pi}{2}\), and local minimum at \(x = \pi\).
Step by step solution
01
Find the First Derivative
To find the relative extrema, we begin by finding the first derivative of the function. The given function is \( f(x) = \sin^2 x \). Using the chain rule, we find the derivative: \( f'(x) = 2 \sin x \cos x = \sin 2x \).
02
Determine Critical Points
Next, determine where the first derivative is zero or undefined to find critical points. Set \( f'(x) = \sin 2x = 0 \). The solutions to this equation within \( 0 < x < 2\pi \) are \( x = \frac{n\pi}{2} \) where \( n \) is an integer. Thus, the critical points in the interval are \( x = \frac{\pi}{2}, \pi, \frac{3\pi}{2} \).
03
Apply the First Derivative Test
Use the critical points to determine whether each point is a local minimum or maximum. Evaluate the sign of \( f'(x) \) on intervals determined by the critical points: - \( (0, \frac{\pi}{2}) \): \( f'(x) > 0 \)- \( (\frac{\pi}{2}, \pi) \): \( f'(x) < 0 \)- \( (\pi, \frac{3\pi}{2}) \): \( f'(x) > 0 \)- \( (\frac{3\pi}{2}, 2\pi) \): \( f'(x) < 0 \)This indicates that \( x = \frac{\pi}{2} \) and \( x = \frac{3\pi}{2} \) are local maxima, and \( x = \pi \) is a local minimum.
04
Find the Second Derivative
Determine the second derivative to confirm the nature of the extrema. The first derivative is \( f'(x) = \sin 2x \). Differentiating again, we find the second derivative: \( f''(x) = 2 \cos 2x \).
05
Apply the Second Derivative Test
Evaluate the second derivative at the critical points:- At \( x = \frac{\pi}{2} \), \( f''(\frac{\pi}{2}) = 2 \cos \pi = -2 \), which is less than 0, confirming a local maximum.- At \( x = \pi \), \( f''(\pi) = 2 \cos(2\pi) = 2 \), which is greater than 0, confirming a local minimum.- At \( x = \frac{3\pi}{2} \), \( f''(\frac{3\pi}{2}) = 2 \cos 3\pi = -2 \), which is less than 0, confirming a local maximum.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative Test
The first derivative test is a critical tool in calculus used to identify relative maxima and minima of a given function. It involves examining the sign changes of the first derivative \( f'(x) \) around critical points:
- **Step 1:** **Find the First Derivative**: This step requires determining the derivative of the function using appropriate differentiation techniques, such as the chain rule or product rule. For example, if you have \( f(x) = \sin^2 x \), its first derivative would be \( f'(x) = \sin 2x \).
- **Step 2:** **Locate Critical Points**: The critical points are determined where the first derivative equals zero or is undefined. For \( \sin 2x \), setting it to zero leads to solutions \( x = \frac{n\pi}{2} \) within a specified range, marking potential extrema.
- **Step 3:** **Apply the Test**: You must evaluate the sign of the derivative in intervals around each critical point. If the derivative changes from positive to negative at a critical point, it's a local maximum. Conversely, changing from negative to positive indicates a local minimum. For instance, at \( x = \frac{\pi}{2} \), if \( f'(x) > 0 \) on one side and \( < 0 \) on the other, you have a maximum.
- **Step 1:** **Find the First Derivative**: This step requires determining the derivative of the function using appropriate differentiation techniques, such as the chain rule or product rule. For example, if you have \( f(x) = \sin^2 x \), its first derivative would be \( f'(x) = \sin 2x \).
- **Step 2:** **Locate Critical Points**: The critical points are determined where the first derivative equals zero or is undefined. For \( \sin 2x \), setting it to zero leads to solutions \( x = \frac{n\pi}{2} \) within a specified range, marking potential extrema.
- **Step 3:** **Apply the Test**: You must evaluate the sign of the derivative in intervals around each critical point. If the derivative changes from positive to negative at a critical point, it's a local maximum. Conversely, changing from negative to positive indicates a local minimum. For instance, at \( x = \frac{\pi}{2} \), if \( f'(x) > 0 \) on one side and \( < 0 \) on the other, you have a maximum.
Second Derivative Test
The second derivative test enhances your analysis by confirming the nature of critical points. Here's how to apply it:
- **Step 1:** **Determine the Second Derivative**: Differentiate the first derivative to find the second derivative \( f''(x) \). If \( f'(x) = \sin 2x \), the second derivative would be \( f''(x) = 2 \cos 2x \).
- **Step 2:** **Evaluate at Critical Points**: Plug the critical points into the second derivative to ascertain the nature of each critical point.
- **Step 1:** **Determine the Second Derivative**: Differentiate the first derivative to find the second derivative \( f''(x) \). If \( f'(x) = \sin 2x \), the second derivative would be \( f''(x) = 2 \cos 2x \).
- **Step 2:** **Evaluate at Critical Points**: Plug the critical points into the second derivative to ascertain the nature of each critical point.
- If \( f''(c) > 0 \), this indicates a local minimum at \( c \).
- If \( f''(c) < 0 \), there is a local maximum at \( c \).
- If \( f''(c) = 0 \), the test is inconclusive, and further investigation is needed.
Critical Points
Critical points are essential for understanding a function's behavior and occur where the derivative equals zero or is undefined. To identify these points, follow these steps:
- **Step 1:** **Calculate First Derivative**: Begin by differentiating the function to find \( f'(x) \).
- **Step 2:** **Solve for Zeroes or Undefined Values**: Setting \( f'(x) = 0 \) helps locate potential critical points. In some cases, the derivative might be undefined at certain points, which can also be critical points.
- **Step 3:** **Verify Within Interval**: Ensure your critical points fall within the specified interval of the problem. For instance, if the interval is between 0 and \( 2\pi \), critical points like \( x = \frac{\pi}{2}, \pi, \text{and} \frac{3\pi}{2} \) are valid.
Identifying critical points paves the way for deeper analysis using subsequent tests, confirming their roles in establishing local maxima or minima.
- **Step 1:** **Calculate First Derivative**: Begin by differentiating the function to find \( f'(x) \).
- **Step 2:** **Solve for Zeroes or Undefined Values**: Setting \( f'(x) = 0 \) helps locate potential critical points. In some cases, the derivative might be undefined at certain points, which can also be critical points.
- **Step 3:** **Verify Within Interval**: Ensure your critical points fall within the specified interval of the problem. For instance, if the interval is between 0 and \( 2\pi \), critical points like \( x = \frac{\pi}{2}, \pi, \text{and} \frac{3\pi}{2} \) are valid.
Identifying critical points paves the way for deeper analysis using subsequent tests, confirming their roles in establishing local maxima or minima.