If a rock is thrown upward on the planet Mathemagicland with a velocity of 12 m/s, its height in meters t seconds later is given by y = 12t- 1.86t2. a. Find the average velocity over the given time intervals. [1, 2] [1, 1.5] [1, 1.01] [1, 1.001] i. ii. ii. iv. b. Estimate the instantaneous velocity when t 1.

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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**Transcription of Educational Material**

---

**1. Problem on Planet Mathemagical:**

A rock is thrown upward on the planet Mathemagicald with a velocity of 12 m/s. Its height in meters \( t \) seconds later is given by the equation:

\[ y = 12t - 1.86t^2 \]

a. **Find the average velocity over the given time intervals.**

   i. \([1, 2]\)

   ii. \([1, 1.5]\)

   iii. \([1, 1.01]\)

   iv. \([1, 1.001]\)

b. **Estimate the instantaneous velocity when \( t = 1 \).**

---

**2. Parabola and Perpendicular Bisectors:**

The figure below shows a point \( P \) on the parabola \( y = -x^2 \) and the point \( Q \) where the perpendicular bisector of \( QP \) intersects the y-axis. As \( P \) approaches the origin along the parabola, what happens to \( Q \)? Does it have a limiting position? If so, find it.

**Explanation of the Diagram:**

- A graph displays a parabola following the equation \( y = -x^2 \).
- Point \( P \) is located on this parabola.
- A line (perpendicular bisector of \( QP \)) intersects the y-axis, determining point \( Q \).

The problem explores the limiting behavior of point \( Q \) as point \( P \) moves toward the origin on the parabola.

---
Transcribed Image Text:**Transcription of Educational Material** --- **1. Problem on Planet Mathemagical:** A rock is thrown upward on the planet Mathemagicald with a velocity of 12 m/s. Its height in meters \( t \) seconds later is given by the equation: \[ y = 12t - 1.86t^2 \] a. **Find the average velocity over the given time intervals.** i. \([1, 2]\) ii. \([1, 1.5]\) iii. \([1, 1.01]\) iv. \([1, 1.001]\) b. **Estimate the instantaneous velocity when \( t = 1 \).** --- **2. Parabola and Perpendicular Bisectors:** The figure below shows a point \( P \) on the parabola \( y = -x^2 \) and the point \( Q \) where the perpendicular bisector of \( QP \) intersects the y-axis. As \( P \) approaches the origin along the parabola, what happens to \( Q \)? Does it have a limiting position? If so, find it. **Explanation of the Diagram:** - A graph displays a parabola following the equation \( y = -x^2 \). - Point \( P \) is located on this parabola. - A line (perpendicular bisector of \( QP \)) intersects the y-axis, determining point \( Q \). The problem explores the limiting behavior of point \( Q \) as point \( P \) moves toward the origin on the parabola. ---
**Problem 3: Graphing a Function**

Sketch a graph of a function \( y = f(x) \) that satisfies all of the following conditions (if possible):

a. \( \lim_{x \to 0^-} f(x) = 1 \)

b. \( \lim_{x \to 0^+} f(x) = -1 \)

c. \( \lim_{x \to 3} f(x) = 0 \)

d. \( f'(3) \) does not exist

e. \( f'(-1) = 2 \)

**Graph Description:**

The graph should demonstrate a function with a jump discontinuity at \( x = 0 \), where the left-hand limit (as \( x \to 0^- \)) approaches 1 and the right-hand limit (as \( x \to 0^+ \)) approaches -1. As \( x \to 3 \), the function should approach 0, but there should be a point or a cusp to indicate that the derivative at \( x = 3 \) does not exist. At \( x = -1 \), the slope of the function, represented by the derivative, should equal 2, indicating a linear tangent.

**Problem 4: Function Limit Exploration**

Is it possible that there exists some function \( g(x) \) and some constant \( L \) such that

\[
\lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = L
\]

but 

\[
\lim_{x \to 2} g(x) 
\]

does not exist? If so, draw a sketch of such a function. If no, explain why not.

**Explanation:**

A function \( g(x) \) can have the left-hand and right-hand limits equal to a constant \( L \), yet the general limit \( \lim_{x \to 2} g(x) \) might not exist if there is a point of discontinuity (such as a hole or a jump at \( x=2 \)). The graph should have left-hand and right-hand limits meeting at \( L \), but a disruption at \( x=2 \).
Transcribed Image Text:**Problem 3: Graphing a Function** Sketch a graph of a function \( y = f(x) \) that satisfies all of the following conditions (if possible): a. \( \lim_{x \to 0^-} f(x) = 1 \) b. \( \lim_{x \to 0^+} f(x) = -1 \) c. \( \lim_{x \to 3} f(x) = 0 \) d. \( f'(3) \) does not exist e. \( f'(-1) = 2 \) **Graph Description:** The graph should demonstrate a function with a jump discontinuity at \( x = 0 \), where the left-hand limit (as \( x \to 0^- \)) approaches 1 and the right-hand limit (as \( x \to 0^+ \)) approaches -1. As \( x \to 3 \), the function should approach 0, but there should be a point or a cusp to indicate that the derivative at \( x = 3 \) does not exist. At \( x = -1 \), the slope of the function, represented by the derivative, should equal 2, indicating a linear tangent. **Problem 4: Function Limit Exploration** Is it possible that there exists some function \( g(x) \) and some constant \( L \) such that \[ \lim_{x \to 2^-} g(x) = \lim_{x \to 2^+} g(x) = L \] but \[ \lim_{x \to 2} g(x) \] does not exist? If so, draw a sketch of such a function. If no, explain why not. **Explanation:** A function \( g(x) \) can have the left-hand and right-hand limits equal to a constant \( L \), yet the general limit \( \lim_{x \to 2} g(x) \) might not exist if there is a point of discontinuity (such as a hole or a jump at \( x=2 \)). The graph should have left-hand and right-hand limits meeting at \( L \), but a disruption at \( x=2 \).
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