Problem 59
Question
a. Find the local extrema of each function on the given interval,
and say where they occur.
b. Graph the function and its derivative together. Comment on
the behavior of \(f\) in relation to the signs and values of \(f^{\prime}. \)
$$
f(x)=\csc ^{2} x-2 \cot x, \quad 0
Step-by-Step Solution
Verified Answer
The local maximum of \( f(x) = \csc^2 x - 2 \cot x \) occurs at \( x = \frac{\pi}{2} \), where \( f(x) = 1 \).
1Step 1: Find the derivative of the function
To find the local extrema, we first need to find the derivative of the function. Given the function: \[ f(x) = \csc^2 x - 2 \cot x \]The derivative \( f'(x) \) can be calculated using the derivatives of cosecant and cotangent:\[ f'(x) = 2 \csc^2 x \cdot \csc x \cdot \cot x + 2 \csc^2 x \cot x \cdot \csc x \]This simplifies to:\[ f'(x) = 2 \csc^2 x \cdot (\csc x \cot x + \csc x \cot x) = 2 \csc^3 x \cdot 2 \cot x \]Thus:\[ f'(x) = 4 \csc^3 x \cot x \]
2Step 2: Set the derivative equal to zero
To find the critical points, we set the derivative equal to zero and solve for \( x \):\[ 4 \csc^3 x \cot x = 0 \]This equation is zero when \( \cot x = 0 \), which occurs when \( x = \frac{\pi}{2} \) within the interval \((0, \pi)\).
3Step 3: Determine the nature of critical points
To determine if \( x = \frac{\pi}{2} \) is a local maximum or minimum, evaluate the sign changes of \( f'(x) \). \( f'(x) \) changes from positive to negative as \( x \) passes through \( \frac{\pi}{2} \), indicating a local maximum at \( x = \frac{\pi}{2} \).
4Step 4: Evaluate the function at the critical point
Substitute the critical point \( x = \frac{\pi}{2} \) back into the original function to find the local extremum value:\[ f\left(\frac{\pi}{2}\right) = \csc^2\left(\frac{\pi}{2}\right) - 2 \cot\left(\frac{\pi}{2}\right) \]\( \csc\left(\frac{\pi}{2}\right) = 1 \) and \( \cot\left(\frac{\pi}{2}\right) = 0 \), so:\[ f\left(\frac{\pi}{2}\right) = 1^2 - 2 \cdot 0 = 1 \]
5Step 5: Comment on the derivative and function graph
Graph \( f(x) \) and its derivative \( f'(x) \). Notice that where \( f'(x) \) is positive, \( f(x) \) is increasing, and where \( f'(x) \) is negative, \( f(x) \) is decreasing.At \( x = \frac{\pi}{2} \), \( f'(x) \) changes sign, confirming a local maximum.
Key Concepts
Critical PointsCosecant FunctionCotangent FunctionDerivative Analysis
Critical Points
Critical points are places where the derivative of a function equals zero or does not exist. These points help us determine where potential local maxima and minima are in a function. When analyzing the function \( f(x) = \csc^2 x - 2 \cot x \), we found its derivative: \( f'(x) = 4 \csc^3 x \cot x \). To find critical points, set \( f'(x) = 0 \). This equation gives us a critical point at \( x = \frac{\pi}{2} \) since \( \cot x = 0 \) at this value.
At the critical point, we can determine whether it is a maximum, minimum, or point of inflection by looking at the behavior of the derivative around this point. If \( f'(x) \) changes from positive to negative, it indicates a local maximum.
In this exercise, at \( x = \frac{\pi}{2} \), \( f'(x) \) changes from positive to negative, confirming it is indeed a local maximum.
At the critical point, we can determine whether it is a maximum, minimum, or point of inflection by looking at the behavior of the derivative around this point. If \( f'(x) \) changes from positive to negative, it indicates a local maximum.
In this exercise, at \( x = \frac{\pi}{2} \), \( f'(x) \) changes from positive to negative, confirming it is indeed a local maximum.
Cosecant Function
The cosecant function, denoted as \( \csc x \), is the reciprocal of the sine function: \( \csc x = \frac{1}{\sin x} \). This function is undefined wherever \( \sin x = 0 \) because division by zero is not possible.
In the given function \( f(x) = \csc^2 x - 2 \cot x \), the term \( \csc^2 x \) denotes the square of \( \csc x \). This introduces additional considerations for undefined points. Notably, \( \csc x \) is undefined at integer multiples of \( \pi \) where the sine function is zero, such as \( 0, \pi, 2\pi, \ldots \), hence these need to be avoided in the problem's context.
Pay attention to these areas outside \( (0, \pi) \) interval boundaries when working with the cosecant function to avoid missteps.
In the given function \( f(x) = \csc^2 x - 2 \cot x \), the term \( \csc^2 x \) denotes the square of \( \csc x \). This introduces additional considerations for undefined points. Notably, \( \csc x \) is undefined at integer multiples of \( \pi \) where the sine function is zero, such as \( 0, \pi, 2\pi, \ldots \), hence these need to be avoided in the problem's context.
Pay attention to these areas outside \( (0, \pi) \) interval boundaries when working with the cosecant function to avoid missteps.
Cotangent Function
The cotangent function, \( \cot x \), is the reciprocal of the tangent function: \( \cot x = \frac{\cos x}{\sin x} \). Like cosecant, it is also undefined when \( \sin x = 0 \) because this results in division by zero. It is crucial to remember these points when dealing with \( \cot x \) in any trigonometric functions.
In the exercise \( f(x) = \csc^2 x - 2 \cot x \), the term \(-2 \cot x\) is critical to understand when seeking critical points. Specifically, it becomes zero when examining where \( \cot x = 0 \), such as at \( x = \frac{\pi}{2} \), helping us identify potential extremum points in the function.
Understanding these undefined points can greatly aid in correctly analyzing and graphing functions involving \( \cot x \).
In the exercise \( f(x) = \csc^2 x - 2 \cot x \), the term \(-2 \cot x\) is critical to understand when seeking critical points. Specifically, it becomes zero when examining where \( \cot x = 0 \), such as at \( x = \frac{\pi}{2} \), helping us identify potential extremum points in the function.
Understanding these undefined points can greatly aid in correctly analyzing and graphing functions involving \( \cot x \).
Derivative Analysis
Derivative analysis is the process of examining the derivative of a function to understand the behavior and trends of the original function. By looking at \( f'(x) \), you can determine where the function is increasing, decreasing, or has a local extremum.
For \( f(x) = \csc^2 x - 2 \cot x \), the derivative \( f'(x) = 4 \csc^3 x \cot x \) helps identify areas where \( f(x) \) changes direction. Setting \( f'(x) = 0 \), the analysis zeroed in on a critical point at \( x = \frac{\pi}{2} \).
Investigating the sign changes around this critical point reveals if \( f'(x) \) changes from positive to negative, indicating a local maximum. Conversely, a change from negative to positive suggests a local minimum.
For \( f(x) = \csc^2 x - 2 \cot x \), the derivative \( f'(x) = 4 \csc^3 x \cot x \) helps identify areas where \( f(x) \) changes direction. Setting \( f'(x) = 0 \), the analysis zeroed in on a critical point at \( x = \frac{\pi}{2} \).
Investigating the sign changes around this critical point reveals if \( f'(x) \) changes from positive to negative, indicating a local maximum. Conversely, a change from negative to positive suggests a local minimum.
- Positive derivative: Function is increasing.
- Negative derivative: Function is decreasing.
- Change in sign of derivative: Potential extremum.
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Problem 59
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