(7) consider the function whose formula is ( f(x) = 3 - 2x ).
- what familiar type of function is ( f )? what can you say about the slope of ( f ) at every value of ( x )?
- compute the average rate of change of ( f ) on the intervals from ( x = 1 ) to ( x = 4 ), from ( x = 3 ) to ( x = 7 ), and from ( x = 5 ) to ( x = 5 + h ); simplify as much as possible. what do you notice?
- use the limit definition of the derivative to compute the exact instantaneous rate of change of ( f ) with respect to ( x ) at the value ( a = 5 ). meaning, compute ( f(5) ) using the limit definition. is your result surprising?
- without doing any additional computations what are the values of ( f(2) ), ( f(pi) ), ( f(-sqrt{2}) )? why?

(8) a water balloon is tossed vertically in the air from a window. the balloon’s height in feet at time ( t ) in seconds after being launched is given by ( s(t) = -16t^2 + 16t + 32 ).
- sketch an accurate, labeled graph of ( s ).
- compute the average rate of change of ( s ) on the time interval ( t = 1 ) to ( t = 2 ). include units and a one sentence explanation of what this value is in context.
- use the limit definition to compute the instantaneous rate of change of ( s ) with respect to time ( t ), at the instant ( a = 1 ). show your work and include proper notation. make sure your answer includes units. write a one sentence explanation of the value you found in context.
- on your graph sketch two lines, one whose slope is your average rate of change of ( s ) from ( t = 1 ) to ( t = 2 ) and the other whose slope is the instantaneous rate of change of ( s ) at the instant ( a = 1 ).

Question

(7) consider the function whose formula is ( f(x) = 3 - 2x ).
- what familiar type of function is ( f )? what can you say about the slope of ( f ) at every value of ( x )?
- compute the average rate of change of ( f ) on the intervals from ( x = 1 ) to ( x = 4 ), from ( x = 3 ) to ( x = 7 ), and from ( x = 5 ) to ( x = 5 + h ); simplify as much as possible. what do you notice?
- use the limit definition of the derivative to compute the exact instantaneous rate of change of ( f ) with respect to ( x ) at the value ( a = 5 ). meaning, compute ( f(5) ) using the limit definition. is your result surprising?
- without doing any additional computations what are the values of ( f(2) ), ( f(pi) ), ( f(-sqrt{2}) )? why?

(8) a water balloon is tossed vertically in the air from a window. the balloon’s height in feet at time ( t ) in seconds after being launched is given by ( s(t) = -16t^2 + 16t + 32 ).
- sketch an accurate, labeled graph of ( s ).
- compute the average rate of change of ( s ) on the time interval ( t = 1 ) to ( t = 2 ). include units and a one sentence explanation of what this value is in context.
- use the limit definition to compute the instantaneous rate of change of ( s ) with respect to time ( t ), at the instant ( a = 1 ). show your work and include proper notation. make sure your answer includes units. write a one sentence explanation of the value you found in context.
- on your graph sketch two lines, one whose slope is your average rate of change of ( s ) from ( t = 1 ) to ( t = 2 ) and the other whose slope is the instantaneous rate of change of ( s ) at the instant ( a = 1 ).

Accepted Answer

### Problem (7)

#### Part 1: Type of Function and Slope
# Explanation:
## Step1: Identify Function Type
The function $ f(x) = 3 - 2x $ is a linear function because it is in the form $ y = mx + b $, where $ m = -2 $ (slope) and $ b = 3 $ (y-intercept).
## Step2: Analyze Slope
For a linear function $ y = mx + b $, the slope $ m $ is constant for all $ x $. So the slope of $ f $ at every $ x $ is $ -2 $.
# Answer:
$ f(x) $ is a linear function. The slope of $ f $ at every $ x $ is constant ($ -2 $).

#### Part 2: Average Rate of Change
The formula for the average rate of change of a function $ f $ on the interval $[a, b]$ is $ \frac{f(b) - f(a)}{b - a} $.

##### Interval $[1, 4]$:
# Explanation:
## Step1: Compute $ f(1) $ and $ f(4) $
$ f(1) = 3 - 2(1) = 1 $
$ f(4) = 3 - 2(4) = -5 $
## Step2: Apply Average Rate Formula
$ \frac{f(4) - f(1)}{4 - 1} = \frac{-5 - 1}{3} = \frac{-6}{3} = -2 $
# Answer:
Average rate of change on $[1, 4]$ is $ -2 $.

##### Interval $[3, 7]$:
# Explanation:
## Step1: Compute $ f(3) $ and $ f(7) $
$ f(3) = 3 - 2(3) = -3 $
$ f(7) = 3 - 2(7) = -11 $
## Step2: Apply Average Rate Formula
$ \frac{f(7) - f(3)}{7 - 3} = \frac{-11 - (-3)}{4} = \frac{-8}{4} = -2 $
# Answer:
Average rate of change on $[3, 7]$ is $ -2 $.

##### Interval $[5, 5 + h]$:
# Explanation:
## Step1: Compute $ f(5) $ and $ f(5 + h) $
$ f(5) = 3 - 2(5) = -7 $
$ f(5 + h) = 3 - 2(5 + h) = 3 - 10 - 2h = -7 - 2h $
## Step2: Apply Average Rate Formula
$ \frac{f(5 + h) - f(5)}{(5 + h) - 5} = \frac{(-7 - 2h) - (-7)}{h} = \frac{-2h}{h} = -2 $ (for $ h
eq 0 $)
# Answer:
Average rate of change on $[5, 5 + h]$ is $ -2 $.

**Observation**: The average rate of change is the same ($ -2 $) for all intervals, which makes sense for a linear function (constant slope).

#### Part 3: Instantaneous Rate of Change at $ a = 5 $ (Limit Definition)
The limit definition of the derivative at $ x = a $ is $ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} $.

# Explanation:
## Step1: Substitute into Limit Definition
$ f'(5) = \lim_{h \to 0} \frac{f(5 + h) - f(5)}{h} $
We already found $ f(5) = -7 $ and $ f(5 + h) = -7 - 2h $ from Part 2.
## Step2: Simplify the Expression
$ \frac{(-7 - 2h) - (-7)}{h} = \frac{-2h}{h} = -2 $ (for $ h
eq 0 $)
## Step3: Take the Limit as $ h \to 0 $
$ \lim_{h \to 0} -2 = -2 $
# Answer:
$ f'(5) = -2 $. This is not surprising because for a linear function, the instantaneous rate of change (derivative) is equal to the slope, which is constant.

#### Part 4: Values of $ f'(2) $, $ f'(\pi) $, $ f'(-\sqrt{2}) $
# Explanation:
For a linear function $ f(x) = mx + b $, the derivative $ f'(x) = m $ (constant) for all $ x $. Since $ f(x) = 3 - 2x $ has slope $ -2 $, the derivative $ f'(x) = -2 $ for any $ x $.
# Answer:
$ f'(2) = -2 $, $ f'(\pi) = -2 $, $ f'(-\sqrt{2}) = -2 $. Because $ f(x) $ is linear, its derivative (instantaneous rate of change) is constant (equal to the slope) for all $ x $.

### Problem (8)

#### Part 1: Sketching the Graph of $ s(t) = -16t^2 + 16t + 32 $
This is a quadratic function (parabola) with $ a = -16 $ (negative, so opens downward), vertex at $ t = -\frac{b}{2a} = -\frac{16}{2(-16)} = 0.5 $. The y-intercept is $ s(0) = 32 $. To sketch:
- Plot the vertex $ (0.5, s(0.5)) $. Compute $ s(0.5) = -16(0.25) + 16(0.5) + 32 = -4 + 8 + 32 = 36 $. So vertex is $ (0.5, 36) $.
- Plot the y-intercept $ (0, 32) $.
- Find the roots (when $ s(t) = 0 $): $ -16t^2 + 16t + 32 = 0 $ → Divide by -16: $ t^2 - t - 2 = 0 $ → Factor: $ (t - 2)(t + 1) = 0 $ → Roots at $ t = 2 $ and $ t = -1 $. Since time can't be negative, we consider $ t = 2 $.
- Draw a downward-opening parabola through these points.

#### Part 2: Average Rate of Change on $[1, 2]$
The formula for average rate of change is $ \frac{s(2) - s(1)}{2 - 1} $.

# Explanation:
## Step1: Compute $ s(1) $ and $ s(2) $
$ s(1) = -16(1)^2 + 16(1) + 32 = -16 + 16 + 32 = 32 $
$ s(2) = -16(4) + 16(2) + 32 = -64 + 32 + 32 = 0 $
## Step2: Apply Average Rate Formula
$ \frac{s(2) - s(1)}{2 - 1} = \frac{0 - 32}{1} = -32 $
# Answer:
The average rate of change is $ -32 $ feet per second. This means the water balloon's height changes by an average of -32 feet per second (it falls 32 feet on average) between $ t = 1 $ and $ t = 2 $ seconds.

#### Part 3: Instantaneous Rate of Change at $ a = 1 $ (Limit Definition)
The limit definition of the derivative at $ t = a $ is $ s'(a) = \lim_{h \to 0} \frac{s(a + h) - s(a)}{h} $.

# Explanation:
## Step1: Substitute $ a = 1 $ into the Limit Definition
$ s'(1) = \lim_{h \to 0} \frac{s(1 + h) - s(1)}{h} $
## Step2: Compute $ s(1 + h) $ and $ s(1) $
$ s(1) = 32 $ (from Part 2)
$ s(1 + h) = -16(1 + h)^2 + 16(1 + h) + 32 $
Expand $ (1 + h)^2 = 1 + 2h + h^2 $:
$ s(1 + h) = -16(1 + 2h + h^2) + 16 + 16h + 32 $
$ = -16 - 32h - 16h^2 + 16 + 16h + 32 $
Simplify: $ (-16 + 16 + 32) + (-32h + 16h) + (-16h^2) = 32 - 16h - 16h^2 $
## Step3: Substitute into the Limit Expression
$ \frac{s(1 + h) - s(1)}{h} = \frac{(32 - 16h - 16h^2) - 32}{h} = \frac{-16h - 16h^2}{h} = \frac{h(-16 - 16h)}{h} = -16 - 16h $ (for $ h
eq 0 $)
## Step4: Take the Limit as $ h \to 0 $
$ \lim_{h \to 0} (-16 - 16h) = -16 $
# Answer:
$ s'(1) = -16 $ feet per second. This means at $ t = 1 $ second, the water balloon's height is changing at a rate of -16 feet per second (it is falling at 16 feet per second at that instant).

#### Part 4: Sketching the Lines
- **Average Rate of Change Line (from $ t = 1 $ to $ t = 2 $)**: This is the secant line connecting the points $ (1, s(1)) = (1, 32) $ and $ (2, s(2)) = (2, 0) $. Its slope is -32 (from Part 2).
- **Instantaneous Rate of Change Line (at $ t = 1 $)**: This is the tangent line to the parabola at $ t = 1 $. Its slope is -16 (from Part 3). To sketch, draw a line at $ t = 1 $ with slope -16, touching the parabola at $ (1, 32) $.

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QUESTION IMAGE

Question

Explanation:

Problem (7)

Part 1: Type of Function and Slope

Step1: Identify Function Type

Step2: Analyze Slope

Step1: Compute \( f(1) \) and \( f(4) \)

Step2: Apply Average Rate Formula

Step1: Compute \( f(3) \) and \( f(7) \)

Step2: Apply Average Rate Formula

Answer:

Part 2: Average Rate of Change

Interval \([1, 4]\):