D 1 The region D above lies between the graphs of y = - 5 – (x – 2)² and y = – 9 +(x – 0)°. It can be described in two ways. 1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of x and provide the interval of x-values that covers the entire region. "top" boundary g2 (x) "bottom" boundary g1(x) interval of x 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 – 6 < y < - 5 the "right" boundary as a piece-wise function f2(y) = For –9 < y < – 6 the "right" boundary f2(y) For -9 < y < – 5 the "left" boundary fi (y)
D 1 The region D above lies between the graphs of y = - 5 – (x – 2)² and y = – 9 +(x – 0)°. It can be described in two ways. 1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of x and provide the interval of x-values that covers the entire region. "top" boundary g2 (x) "bottom" boundary g1(x) interval of x 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 – 6 < y < - 5 the "right" boundary as a piece-wise function f2(y) = For –9 < y < – 6 the "right" boundary f2(y) For -9 < y < – 5 the "left" boundary fi (y)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:D
The region D above lies between the graphs of y
- 5 - (x – 2) and y =
- 9 +
(r – 0)°. It can
|
be described in two ways.
1. If we visualize the region having "top" and "bottom" boundaries, express each as functions of x and
provide the interval of x-values that covers the entire region.
"top" boundary g2 (x)
=
"bottom" boundary g1(x) =
interval of x 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 – 6 <y < – 5 the "right" boundary as a piece-wise function f2(y)
For – 9 < y < – 6 the "right" boundary f2(y)
For – 9 < y < – 5 the "left" boundary fi (y) =
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