5) Sketch the graphs of the functions f and g. f(x) = x2, g(x) = 1/x2 Find the area of the region enclosed by these graphs and the vertical lines x = 1 and x = 3. = ? square units
5) Sketch the graphs of the functions f and g. f(x) = x2, g(x) = 1/x2 Find the area of the region enclosed by these graphs and the vertical lines x = 1 and x = 3. = ? square units
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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5)
Sketch the graphs of the functions f and g.
f(x) = x2, g(x) = 1/x2
Find the area of the region enclosed by these graphs and the vertical lines
x = 1 and x = 3.

Transcribed Image Text:The image contains four graphs comparing the functions \( f(x) = x^2 \) and \( g(x) = \frac{1}{x^2} \), depicted on a Cartesian plane with \( x \)- and \( y \)-coordinates ranging from \(-10\) to \(10\).
---
### Graph Descriptions:
1. **Top Left Graph:**
- **Red Curve (f(x) = x^2):** This is a parabola opening upwards, crossing the y-axis at the origin (0,0).
- **Blue Curve (g(x) = 1/x^2):** This hyperbola opens upwards and downwards, approaching the x-axis as an asymptote. It remains undefined at \( x = 0 \) and exhibits vertical asymptotic behavior.
2. **Top Right Graph:**
- Red Curve (g(x) = -1/x^2): This hyperbola opens downwards, similar in shape to the inverse square but negatively mirrored.
- Blue Curve (f(x) = x^2): Same parabola as in the first graph.
3. **Bottom Left Graph:**
- **Red Curve (-1/x^2):** Part of this curve appears in the first quadrant, the rest in the second, mirroring the traditional hyperbola horizontally.
- **Blue Curve (x^2):** Parabola with minimal visible adjustment in position.
4. **Bottom Right Graph:**
- **Red Curve (1/x^2):** Identical to the first graph’s hyperbola behavior in shape, appearing in both the first and second quadrants.
- **Blue Curve (x^2):** Parabola shown just as it appears in previous graphs.
### Observations:
- **Domain and Range:**
- The domain of \( f(x) = x^2 \) includes all real numbers, while \( g(x) = \frac{1}{x^2} \) excludes \( x = 0 \).
- The range of \( f(x) = x^2 \) is \( y \geq 0 \); \( g(x) = \frac{1}{x^2} \) has \( y > 0 \).
- **Asymptotic Behavior:**
- \( g(x) = \frac{1}{x^2} \) tends towards infinity as \( x \
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