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Magazine Chapter 7: Problem 36
Exer. 33-42: Sketch the graph of the equation. $$ y=\sin ^{-1}(x-2)+\frac{\pi}{2} $$
Short Answer
Expert verified
Graph involves a domain shift and a vertical shift to inverse sine, from \(x=1\) to \(x=3\).
Step by step solution
01
Understand the Function
The given function is \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \), which involves an inverse sine function. The function \( \sin^{-1}(x) \) has a domain of \([-1, 1]\) and a range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\). Generally, it represents the angle whose sine is \(x\).
02
Determine the Domain
For the function \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \), solve for the domain by setting the expression inside the inverse sine function to be between -1 and 1: \( -1 \leq x-2 \leq 1 \). By solving this inequality:- Add 2 to all parts: \( -1 + 2 \leq x-2 + 2 \leq 1 + 2 \)- This simplifies to: \( 1 \leq x \leq 3 \). Thus, the domain is \( [1, 3] \).
03
Identify Key Points
For sketching, identify key points by choosing values of \(x\) within the domain. Calculate corresponding \(y\) values:- For \(x = 1\): \( y = \sin^{-1}(1-2) + \frac{\pi}{2} = \sin^{-1}(-1) + \frac{\pi}{2} = -\frac{\pi}{2} + \frac{\pi}{2} = 0 \)- For \(x = 2\): \( y = \sin^{-1}(2-2) + \frac{\pi}{2} = \sin^{-1}(0) + \frac{\pi}{2} = 0 + \frac{\pi}{2} = \frac{\pi}{2} \)- For \(x = 3\): \( y = \sin^{-1}(3-2) + \frac{\pi}{2} = \sin^{-1}(1) + \frac{\pi}{2} = \frac{\pi}{2} + \frac{\pi}{2} = \pi \)
04
Sketch the Graph
Plot the points from Step 3: \((1, 0), (2, \frac{\pi}{2}), (3, \pi)\). Since these values reflect the range for \( \sin^{-1}(x-2) \) shifted upwards by \(\frac{\pi}{2}\), draw a smooth curve that starts at \((1, 0)\), passes through \((2, \frac{\pi}{2})\), and ends at \((3, \pi)\). The curve will be smooth and increasing.
05
Consider Graph Transformation
The graph of \( y = \sin^{-1}(x-2)\) is shifted up by \( \frac{\pi}{2} \). This vertical translation means each point on the inverse sine curve will move upwards by \( \frac{\pi}{2} \). This impacts the y-values, keeping the shape of the inverse sine but adjusting locations.
06
Label the Graph
Label the x-axis with domain boundaries (1 to 3) and y-axis with key values (0 to \(\pi\)) at the calculated points. Clearly mark the curve and points \((1,0)\), \((2, \frac{\pi}{2})\), and \((3, \pi)\) on the graph.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Inverse Sine Function
The inverse sine function, denoted as \( \sin^{-1}(x) \) or \( \arcsin(x) \), is a fundamental concept in trigonometry. It essentially answers the question: "What angle has a sine of \( x \)?" This function is particularly useful for finding angles when the sine of those angles is known.
Here's what you should remember about the inverse sine function:
Here's what you should remember about the inverse sine function:
- It yields angles measured in radians.
- The standard range of \( \sin^{-1}(x) \) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), meaning the output is limited to these angles.
- Its domain, or the set of possible input values for \( x \), is \([-1, 1]\) since the sine of an angle can only range from -1 to 1.
- For example, \( \sin^{-1}(-0.5) \) will give you an angle within \([-\frac{\pi}{2}, \frac{\pi}{2}]\) whose sine is -0.5.
Domain and Range
Understanding the domain and range of a function is crucial for correctly sketching its graph. The *domain* refers to all possible input values, while the *range* covers all possible output values the function can produce.
For the function \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \):
Thus, the shift in the range is due to the constant addition, which results in elevating each y-value by \(\frac{\pi}{2}\).
For the function \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \):
- The expression \( x-2 \) inside the inverse sine requires that \(-1 \leq x-2 \leq 1\). Solving this inequality leads to a domain of \([1, 3]\).
- This domain tells us that \( x \) values will range from 1 to 3.
- The range of the function \( \sin^{-1}(x-2) \) alone is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), but adding \(\frac{\pi}{2}\) shifts this to \([0, \pi]\).
Thus, the shift in the range is due to the constant addition, which results in elevating each y-value by \(\frac{\pi}{2}\).
Vertical Translation of Functions
In the context of transformations, a vertical translation refers to moving a graph up or down along the y-axis. This occurs when a constant is added or subtracted to the function's values.
For the function \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \):
Vertical translations like this do not affect the horizontal position of the graph (x-values remain unaffected), only the vertical positioning (y-values are increased by the shift amount).
For the function \( y = \sin^{-1}(x-2) + \frac{\pi}{2} \):
- The term \(+ \frac{\pi}{2}\) causes a vertical shift upwards by \(\frac{\pi}{2}\).
- This means every point on the graph is translated up, preserving the shape of the graph but elevating it in its entirety.
- The effect of this is to move the range of the original inverse sine function, from \([-\frac{\pi}{2}, \frac{\pi}{2}]\) to \([0, \pi]\), reflecting this upward movement on the y-axis.
Vertical translations like this do not affect the horizontal position of the graph (x-values remain unaffected), only the vertical positioning (y-values are increased by the shift amount).
