Problem 133
Question
Use your graphing utility. Graph \(f(x)=\sin ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in relation to the signs and values of \(f^{\prime}\) and \(f^{\prime \prime}\)
Step-by-Step Solution
Verified Answer
The graph of \( f(x) = \sin^{-1}(x) \) is an increasing curve with an inflection point at \( x = 0 \).
1Step 1: Graph the Function
First, we need to graph the function \( f(x) = \sin^{-1}(x) \). This function is the inverse sine function, which means it will be defined for \(-1 \leq x \leq 1\) and \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).
2Step 2: Graph the First Derivative
Now, find and graph the first derivative of \( f(x) \). The derivative of \( f(x) = \sin^{-1}(x) \) is \( f'(x) = \frac{1}{\sqrt{1-x^2}} \). This derivative is undefined at \( x = 1 \) and \( x = -1 \) and remains positive in the interval \( (-1, 1) \), indicating an increasing function.
3Step 3: Graph the Second Derivative
The second derivative \( f''(x) \) can be found by differentiating \( f'(x) = \frac{1}{\sqrt{1-x^2}} \) again. The result is \( f''(x) = \frac{x}{(1-x^2)^{3/2}} \). This second derivative changes sign at \( x = 0 \): it is negative for \( x < 0 \) and positive for \( x > 0 \), indicating a point of inflection.
4Step 4: Combine Graphs and Interpret Behavior
When observing the graph of \( f(x) \), notice that it is a smooth, increasing curve from \(-1\) to \(1\). The first derivative, \( f'(x) \), ensures \( f(x) \) is increasing, as it is positive across its domain. The second derivative, \( f''(x) \), confirms a concave down shape before \(x = 0\) and a concave up shape after, indicating an inflection point at \( x = 0 \).
Key Concepts
Derivative ConceptsConcavity and Inflection PointsGraphing Techniques
Derivative Concepts
Understanding derivatives is crucial when analyzing any function, especially inverse trigonometric functions like the inverse sine function, denoted as \( f(x) = \sin^{-1}(x) \). The first derivative of a function, \( f'(x) \), tells us about the rate of change or the slope of the function at any given point. For \( f(x) = \sin^{-1}(x) \), the first derivative is given by \( f'(x) = \frac{1}{\sqrt{1-x^2}} \).
This derivative is defined for \(-1 < x < 1\) and clearly indicates that the slope is always positive within this interval. Hence, the function is always increasing on this interval. However, at \( x = 1 \) and \( x = -1 \), the derivative is undefined, which corresponds to vertical tangents or endpoints on the graph.
Examining the first derivative provides insight into the overall trend of the graph: if \( f'(x) > 0 \), the function is increasing, and if \( f'(x) < 0 \), it would be decreasing. In this case, with \( f'(x) \) positive, \( f(x) = \sin^{-1}(x) \) rises consistently across its domain.
This derivative is defined for \(-1 < x < 1\) and clearly indicates that the slope is always positive within this interval. Hence, the function is always increasing on this interval. However, at \( x = 1 \) and \( x = -1 \), the derivative is undefined, which corresponds to vertical tangents or endpoints on the graph.
Examining the first derivative provides insight into the overall trend of the graph: if \( f'(x) > 0 \), the function is increasing, and if \( f'(x) < 0 \), it would be decreasing. In this case, with \( f'(x) \) positive, \( f(x) = \sin^{-1}(x) \) rises consistently across its domain.
Concavity and Inflection Points
Concavity describes the direction a curve opens. For the inverse sine function, the second derivative \( f''(x) = \frac{x}{(1-x^2)^{3/2}} \) helps us understand this concept.
Concavity is determined by the sign of the second derivative:
At an inflection point, the second derivative is typically zero or undefined, but more importantly, the graph changes its curvature. Thus, \( x = 0 \) denotes a switch in the shape of \( f(x) = \sin^{-1}(x) \), marking a crucial transition.
Concavity is determined by the sign of the second derivative:
- If \( f''(x) > 0 \), the graph is concave up (like a smile).
- If \( f''(x) < 0 \), the graph is concave down (like a frown).
At an inflection point, the second derivative is typically zero or undefined, but more importantly, the graph changes its curvature. Thus, \( x = 0 \) denotes a switch in the shape of \( f(x) = \sin^{-1}(x) \), marking a crucial transition.
Graphing Techniques
To accurately visualize \( f(x) = \sin^{-1}(x) \) and its derivatives, graphing techniques help highlight the behavior and traits of the function. Start with the basic graph of the inverse sine function. This curve is smooth, resembling a flattened S-shape.
Next, plot the first derivative \( f'(x) \), which is positive for \(-1 < x < 1\), confirming that the original function is increasing throughout the interval. The derivative graph rises steeply as \( x \) approaches the endpoints, hinting at the limitations in its domain.
Combining these graphs provides a clear picture of how \( f(x) = \sin^{-1}(x) \) behaves. This technique reinforces the impact of derivatives on graph interpretation, making it easier to visualize critical features such as slope, concavity, and inflection points.
Next, plot the first derivative \( f'(x) \), which is positive for \(-1 < x < 1\), confirming that the original function is increasing throughout the interval. The derivative graph rises steeply as \( x \) approaches the endpoints, hinting at the limitations in its domain.
- First derivative: confirms the increasing nature.
- Second derivative: elucidates changes in concavity.
Combining these graphs provides a clear picture of how \( f(x) = \sin^{-1}(x) \) behaves. This technique reinforces the impact of derivatives on graph interpretation, making it easier to visualize critical features such as slope, concavity, and inflection points.
Other exercises in this chapter
Problem 132
Find the length of each curve. $$y=\ln (\csc x) \text { from } x=\pi / 6 \text { to } x=\pi / 4$$
View solution Problem 133
a. Show that \(\int \ln x d x=x \ln x-x+C\) b. Find the average value of \(\ln x\) over \([1, e]\)
View solution Problem 134
Find the average value of \(f(x)=1 / x\) on [1,2].
View solution Problem 134
Use your graphing utility. Graph \(f(x)=\tan ^{-1} x\) together with its first two derivatives. Comment on the behavior of \(f\) and the shape of its graph in r
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