The region D above lies between the graphs of y = 1 - (x - 1)² and y describe in two ways. - 3+ -(x + 1)³. It can be

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Chapter2: Second-order Linear Odes
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Question
D
D
Q
The region D above lies between the graphs of y = 1 - (x - 1)² and y
describe in two ways.
- 3+
7/
(x + 1)³. It can be
1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of a and
provide the interval of -values that covers the entire region.
"top" boundary 9₂(x) =
"bottom" boundary 9₁(x) =
interval of values that covers the region =
2. If we visualize the region having "right" and "left" boundaries, then the "right" boundary must be defined
piece-wise. Express each as functions of y for the provided intervals of y-values that covers the entire
region.
For 0 ≤ y ≤ 1 the "right" boundary as a piece-wise function f2(y) =
For -3 < y < 0 the "right" boundary f2(y) =
For -3 < y < 1 the "left" boundary f₁(y) =
Transcribed Image Text:D D Q The region D above lies between the graphs of y = 1 - (x - 1)² and y describe in two ways. - 3+ 7/ (x + 1)³. It can be 1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of a and provide the interval of -values that covers the entire region. "top" boundary 9₂(x) = "bottom" boundary 9₁(x) = interval of values that covers the region = 2. If we visualize the region having "right" and "left" boundaries, then the "right" boundary must be defined piece-wise. Express each as functions of y for the provided intervals of y-values that covers the entire region. For 0 ≤ y ≤ 1 the "right" boundary as a piece-wise function f2(y) = For -3 < y < 0 the "right" boundary f2(y) = For -3 < y < 1 the "left" boundary f₁(y) =
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