Problem 30
Question
Finding Points of Inflection In Exercises \(15-30\) , find the points of inflection and discuss the concavity of the graph of the function. $$ f(x)=x+2 \cos x, \quad[0,2 \pi] $$
Step-by-Step Solution
Verified Answer
The points of inflection are \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). In the intervals \((0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\), the graph of the function is concave down, up, down, respectively.
1Step 1: Find the first derivative of the function
To begin with, we need to find the first derivative of \(f(x)\). The function \(f(x)=x+2 \cos x\). By using the chain rule, its first derivative \(f'(x)\) is given by \(f'(x)=1-2 \sin x\).
2Step 2: Find the second derivative of the function
Next, we need to find the second derivative of \(f(x)\), \(f''(x)\), which is the derivative of the first derivative. So, \(f''(x)=-2 \cos x\).
3Step 3: Find the potential points of inflection
The potential points of inflection can be found by equating the second derivative to zero and solving for \(x\), i.e. solving the equation: -2 cos(x) = 0. The solutions are \(x= \frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). However, since the interval is \([0,2\pi]\), the points of inflection are the solutions that lie within this interval.
4Step 4: Confirm the points of inflection and discuss the concavity of the function
We test the sign of the second derivative in the intervals \((0, \frac{\pi}{2})\), \((\frac{\pi}{2}, \frac{3\pi}{2})\) and \((\frac{3\pi}{2}, 2\pi)\) and find that the signs are negative, positive, and negative, respectively. So, the function is concave down, up, down in these intervals. Thus the points of inflection are indeed those found in Step 3.
Key Concepts
Concavity of a GraphSecond Derivative TestCalculus of a Single Variable
Concavity of a Graph
The concavity of a graph reveals how the shape of a function's curve bends over its domain. To visualize this, imagine holding a graph from below; if the curve bends towards you, it is concave up. Similarly, if it bends away from you, it is concave down.
Mathematically, the second derivative of a function, denoted as \( f''(x) \), helps us determine concavity. If \( f''(x) > 0 \) at a certain interval, the function is concave up in that interval. Conversely, if \( f''(x) < 0 \) in an interval, the function is concave down. In the exercise, finding where \( f''(x) \) changes sign indicates potential points of inflection, which are key points where the graph changes concavity.
Mathematically, the second derivative of a function, denoted as \( f''(x) \), helps us determine concavity. If \( f''(x) > 0 \) at a certain interval, the function is concave up in that interval. Conversely, if \( f''(x) < 0 \) in an interval, the function is concave down. In the exercise, finding where \( f''(x) \) changes sign indicates potential points of inflection, which are key points where the graph changes concavity.
Second Derivative Test
The second derivative test is a practical tool in calculus used to analyze the concavity of a graph and identify local maxima and minima. It involves taking the derivative of the first derivative, which results in the second derivative of the function, \( f''(x) \).
For a critical point \( x \) where \( f'(x) = 0 \), if \( f''(x) > 0 \), this means \( x \) is a local minimum. If \( f''(x) < 0 \), \( x \) is a local maximum. However, if \( f''(x) = 0 \), the second derivative test is inconclusive, and other methods, such as the first derivative test, may be used to classify the critical point. In our exercise, after finding the first derivative, we applied the second derivative test to pinpoint potential inflection points where the concavity potentially changes.
For a critical point \( x \) where \( f'(x) = 0 \), if \( f''(x) > 0 \), this means \( x \) is a local minimum. If \( f''(x) < 0 \), \( x \) is a local maximum. However, if \( f''(x) = 0 \), the second derivative test is inconclusive, and other methods, such as the first derivative test, may be used to classify the critical point. In our exercise, after finding the first derivative, we applied the second derivative test to pinpoint potential inflection points where the concavity potentially changes.
Calculus of a Single Variable
Calculus of a single variable focuses on functions that involve only one independent variable. This branch of calculus deals with the concepts of derivatives and integrals of these functions. In understanding the behavior of single-variable functions, like \( f(x) = x + 2 \cos x \), we use differentiation to find rates of change and curvature, and integration to find areas under curves.
In terms of performing differentiation on \( f(x) \), as in the exercise provided, it's crucial to follow a systematic process. First, find the derivative of the function, \( f'(x) \), then proceed to find the second derivative, \( f''(x) \), which is the derivative of \( f'(x) \). These derivatives are essential in understanding the function's behavior, identifying its critical points, points of inflection, and overall concavity within a given interval.
In terms of performing differentiation on \( f(x) \), as in the exercise provided, it's crucial to follow a systematic process. First, find the derivative of the function, \( f'(x) \), then proceed to find the second derivative, \( f''(x) \), which is the derivative of \( f'(x) \). These derivatives are essential in understanding the function's behavior, identifying its critical points, points of inflection, and overall concavity within a given interval.
Other exercises in this chapter
Problem 30
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Applying the First Derivative Test In Exercises \(17-40\) , (a) find the critical numbers of \(f\) (if any), (b) find the open interval(s) on which the function
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