The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-1, 2]. Vertical cross sections that are perpendicular to the x-axis are squares. Algebraically, find the area of each square. 6 3-2-1 1 2 3 4 5 6 7 = Y -2+ 3-
The base of a three-dimensional figure is bound by the line y = 4 - x on the interval [-1, 2]. Vertical cross sections that are perpendicular to the x-axis are squares. Algebraically, find the area of each square. 6 3-2-1 1 2 3 4 5 6 7 = Y -2+ 3-
Calculus: Early Transcendentals
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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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![The base of a three-dimensional figure is bound by the line \( y = 4 - x \) on the interval \([-1, 2]\). Vertical cross sections that are perpendicular to the x-axis are squares. Algebraically, find the area of each square.
**Diagram Explanation:**
The diagram shows a graph with the x-axis and y-axis. The line \( y = 4 - x \) is plotted, creating a right triangle in the positive section of the graph.
- ***X-axis Interval:*** The line is bounded on the interval \([-1, 2]\).
- ***Y-axis Intersection:*** The line crosses the y-axis at \( y = 4 \).
- ***Shaded Region:*** The region below the line from \( x = -1 \) to \( x = 2 \) represents the base of the three-dimensional figure, highlighting the triangular area that corresponds to the base.
To find the area of each square (the cross sections perpendicular to the x-axis), use the formula for the area of a square, \( \text{Area} = \text{side}^2 \).
Given:
- The side length of each square is equal to \( y = 4 - x \).
So the area \( A(x) \) of each square is:
\[ A(x) = (4 - x)^2 \]
This expression represents the area of the squares for values of \( x \) in the interval \([-1, 2]\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa18a6723-a2ab-43ae-9c9a-35233fc8cb89%2Fb5b4e622-67d4-48d5-b83e-93947a70a07f%2Fxu81wdd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The base of a three-dimensional figure is bound by the line \( y = 4 - x \) on the interval \([-1, 2]\). Vertical cross sections that are perpendicular to the x-axis are squares. Algebraically, find the area of each square.
**Diagram Explanation:**
The diagram shows a graph with the x-axis and y-axis. The line \( y = 4 - x \) is plotted, creating a right triangle in the positive section of the graph.
- ***X-axis Interval:*** The line is bounded on the interval \([-1, 2]\).
- ***Y-axis Intersection:*** The line crosses the y-axis at \( y = 4 \).
- ***Shaded Region:*** The region below the line from \( x = -1 \) to \( x = 2 \) represents the base of the three-dimensional figure, highlighting the triangular area that corresponds to the base.
To find the area of each square (the cross sections perpendicular to the x-axis), use the formula for the area of a square, \( \text{Area} = \text{side}^2 \).
Given:
- The side length of each square is equal to \( y = 4 - x \).
So the area \( A(x) \) of each square is:
\[ A(x) = (4 - x)^2 \]
This expression represents the area of the squares for values of \( x \) in the interval \([-1, 2]\).
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