The region D above lies between the graphs of y = - 5 – (x – 2) and 1 9 + -(x – 0)°. It can be described in two ways. 9 y = 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) = | -5 – (x – 2)² of "bottom" boundary g1(x) = | -9+ (x – 0)3 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 – 6 < y < - 5 the "right" boundary as a piece-wise function f2(y) = For – 9 < y < – 6 the "right" boundary f2(y) = %3D For – 9 < y < 5 the "left" boundary f1(y) =
The region D above lies between the graphs of y = - 5 – (x – 2) and 1 9 + -(x – 0)°. It can be described in two ways. 9 y = 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) = | -5 – (x – 2)² of "bottom" boundary g1(x) = | -9+ (x – 0)3 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 – 6 < y < - 5 the "right" boundary as a piece-wise function f2(y) = For – 9 < y < – 6 the "right" boundary f2(y) = %3D For – 9 < y < 5 the "left" boundary f1(y) =
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:The region D above lies between the graphs of y =
– 5 – (x – 2)² and
1
9 + -(x – 0)°. It can be described in two ways.
3
y =
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) = | -5 – (x – 2)²
"bottom" boundary g1(x)
(x – 0)3
-9 +
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 f1(y) =
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