Problem 21
Question
For the functions in problems, do the following: (a) Find \(f^{\prime}\) and \(f^{\prime \prime}\). (b) Find the critical points of \(f\). (c) Find any inflection points of \(f\). (d) Evaluate \(f\) at its critical points and at the endpoints of the given interval. Identify local and global maxima and minima of \(f\) in the interval. (e) Graph \(f\). $$ f(x)=x+\sin x \quad(0 \leq x \leq 2 \pi) $$
Step-by-Step Solution
Verified Answer
Critical point at \(x=\pi\), inflection points at \(x=0, \pi, 2\pi\). Global min: \(f(0)=0\), global max: \(f(2\pi)=2\pi\).
1Step 1: Find the First Derivative
To find the first derivative of the function \(f(x) = x + \sin x\), we use the basic derivative rules \(\frac{d}{dx}(x) = 1\) and \(\frac{d}{dx}(\sin x) = \cos x\). Thus, \(f^{\prime}(x) = 1 + \cos x\).
2Step 2: Find the Second Derivative
Differentiate \(f^{\prime}(x) = 1 + \cos x\) to find the second derivative. We use \(\frac{d}{dx}(\cos x) = -\sin x\), hence \(f^{\prime \prime}(x) = -\sin x\).
3Step 3: Find the Critical Points
Critical points occur where \(f^{\prime}(x) = 0\). Solve \(1 + \cos x = 0\). This implies \(\cos x = -1\), which occurs at \(x = \pi\) in the interval \([0, 2\pi]\).
4Step 4: Find Inflection Points
Inflection points occur where \(f^{\prime \prime}(x) = 0\) or changes sign. Solve \(-\sin x = 0\). This equation is satisfied when \(\sin x = 0\). In the interval, this occurs at \(x = 0, \pi, 2\pi\). We check sign changes in \(f^{\prime \prime}(x)\) around these points.
5Step 5: Evaluate Function at Critical and Boundary Points
Evaluate \(f(x) = x + \sin x\) at critical and endpoint values. Calculate \(f(0) = 0 + \sin(0) = 0\), \(f(2\pi) = 2\pi + \sin(2\pi) = 2\pi\), and \(f(\pi) = \pi + \sin(\pi) = \pi\).
6Step 6: Determine Local and Global Extrema
Compare values: \(f(0) = 0\), \(f(\pi) = \pi\), and \(f(2\pi) = 2\pi\). The least value is at \(x=0\) (local and global minimum), and the greatest value is at \(x=2\pi\) (local and global maximum).
7Step 7: Sketch the Graph of \(f(x)\)
Sketch \(f(x) = x + \sin x\). At \(x = 0\), the graph starts at \(0\). At \(x = \pi\), the graph passes through \(\pi\). Finally, at \(x = 2\pi\), it reaches \(2\pi\). The curve is smooth with slight oscillations due to \(\sin x\).
Key Concepts
Understanding Critical Points in CalculusExploring Inflection PointsThe Art of Sketching Graphs
Understanding Critical Points in Calculus
Critical points in calculus are where the derivative of a function is equal to zero or undefined, indicating potential maxima or minima of the function. By finding these points, you can determine where a graph could change direction from increasing to decreasing, or vice versa.
In our example, for the function \(f(x) = x + \sin x\), the derivative is \(f'(x) = 1 + \cos x\). We set this equal to zero to find critical points: \(1 + \cos x = 0\). This equation simplifies to \(\cos x = -1\).
Solving \(\cos x = -1\) within the interval \([0, 2\pi]\), we find that \(x = \pi\) is a critical point. This is a crucial step in understanding the behavior of the function, as it helps us identify where the curve might exhibit a local extremum.
To properly classify these points as maxima, minima, or points of inflection, other methods such as the second derivative test or examining the function's behavior around these points may be applied.
In our example, for the function \(f(x) = x + \sin x\), the derivative is \(f'(x) = 1 + \cos x\). We set this equal to zero to find critical points: \(1 + \cos x = 0\). This equation simplifies to \(\cos x = -1\).
Solving \(\cos x = -1\) within the interval \([0, 2\pi]\), we find that \(x = \pi\) is a critical point. This is a crucial step in understanding the behavior of the function, as it helps us identify where the curve might exhibit a local extremum.
To properly classify these points as maxima, minima, or points of inflection, other methods such as the second derivative test or examining the function's behavior around these points may be applied.
Exploring Inflection Points
Inflection points are where a curve changes its concavity; this means the graph changes from curving upwards to downwards or vice versa. This is identified by the second derivative of the function. If the second derivative changes sign at a point, that point is an inflection point.
For \(f(x) = x + \sin x\), we find that the second derivative is \(f''(x) = -\sin x\). To find inflection points, set this derivative equal to zero: \(-\sin x = 0\), which simplifies to \(\sin x = 0\).
Within the interval \([0, 2\pi]\), \(\sin x = 0\) at \(x = 0, \pi, 2\pi\). We consider these points and check the behavior of \(f''(x)\) around them to confirm any sign changes, signifying inflection points. For each \(x\), if \(f''(x)\) transitions from negative to positive or positive to negative, then \(x\) is indeed an inflection point.
For \(f(x) = x + \sin x\), we find that the second derivative is \(f''(x) = -\sin x\). To find inflection points, set this derivative equal to zero: \(-\sin x = 0\), which simplifies to \(\sin x = 0\).
Within the interval \([0, 2\pi]\), \(\sin x = 0\) at \(x = 0, \pi, 2\pi\). We consider these points and check the behavior of \(f''(x)\) around them to confirm any sign changes, signifying inflection points. For each \(x\), if \(f''(x)\) transitions from negative to positive or positive to negative, then \(x\) is indeed an inflection point.
The Art of Sketching Graphs
Sketching the graph involves understanding the nature of the function together with any known critical and inflection points. Here, we want to visualize \(f(x) = x + \sin x\) over \([0, 2\pi]\).
Begin by noting the key points: critical points where the function potentially changes direction, inflection points, and behavior at the edges of the interval. We know \(x = \pi\) is a critical point and potential extremum. The graph passes through \((0, 0)\), \((\pi, \pi)\), and \((2\pi, 2\pi)\).
The curve itself is influenced by the oscillating nature of \(\sin x\). Its presence causes small oscillations, adding nuance to the otherwise linear \(y = x\). Visualize these slight waves as the graph transitions through the critical and inflection points.
To ensure accuracy:
Begin by noting the key points: critical points where the function potentially changes direction, inflection points, and behavior at the edges of the interval. We know \(x = \pi\) is a critical point and potential extremum. The graph passes through \((0, 0)\), \((\pi, \pi)\), and \((2\pi, 2\pi)\).
The curve itself is influenced by the oscillating nature of \(\sin x\). Its presence causes small oscillations, adding nuance to the otherwise linear \(y = x\). Visualize these slight waves as the graph transitions through the critical and inflection points.
To ensure accuracy:
- Check the function at critical and endpoint values.
- Note the identified extrema: local and global minima or maxima.
- Observe how the function behaves as it approaches these key points.
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