math 122 - activity 3
1. let ( f(x) = 3x^2 - x ).
(a) find the equation of the secant line from ( x = 1 ) to ( x = 2 ).
(b) find the equation of the tangent line at ( x = 3 ). (find the slope of the tangent line by taking a limit. do not using differentiation formulas.)

2. if the tangent line to ( y = f(x) ) at ( (2, 1) ) passes through the point ( (5, 13) ), find ( f(2) ) and ( f(2) ).

3. an object is dropped from the top of a building 256 ft tall. its position (in ft) at time ( t ) seconds is given by ( f(t) = 256 - 16t^2 ).
(a) find its average velocity from time ( t = 0 ) to the time it hits the ground.
(b) find its (instantaneous) velocity when it hits the ground by taking a limit. (do not use differentiation formulas.)

Question

math 122 - activity 3
1. let ( f(x) = 3x^2 - x ).
(a) find the equation of the secant line from ( x = 1 ) to ( x = 2 ).
(b) find the equation of the tangent line at ( x = 3 ). (find the slope of the tangent line by taking a limit. do not using differentiation formulas.)

2. if the tangent line to ( y = f(x) ) at ( (2, 1) ) passes through the point ( (5, 13) ), find ( f(2) ) and ( f(2) ).

3. an object is dropped from the top of a building 256 ft tall. its position (in ft) at time ( t ) seconds is given by ( f(t) = 256 - 16t^2 ).
(a) find its average velocity from time ( t = 0 ) to the time it hits the ground.
(b) find its (instantaneous) velocity when it hits the ground by taking a limit. (do not use differentiation formulas.)

Accepted Answer

### Problem 1(a)
# Explanation:
## Step 1: Find $ f(1) $ and $ f(2) $
For $ f(x) = 3x^2 - x $, when $ x = 1 $: $ f(1)=3(1)^2 - 1 = 3 - 1 = 2 $. When $ x = 2 $: $ f(2)=3(2)^2 - 2 = 12 - 2 = 10 $.
## Step 2: Calculate the slope of the secant line
The slope $ m $ of the secant line between $ x = 1 $ and $ x = 2 $ is $ m=\frac{f(2)-f(1)}{2 - 1}=\frac{10 - 2}{1}=8 $.
## Step 3: Use point - slope form to find the equation
Using the point - slope form $ y - y_1=m(x - x_1) $ with the point $ (1,f(1))=(1,2) $ and $ m = 8 $, we get $ y - 2=8(x - 1) $, which simplifies to $ y=8x - 8 + 2=8x - 6 $.
# Answer: The equation of the secant line is $ y = 8x-6 $

### Problem 1(b)
# Explanation:
## Step 1: Recall the definition of the slope of the tangent line
The slope of the tangent line at $ x = 3 $ is given by $ m=\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h} $.
First, find $ f(3)=3(3)^2-3=27 - 3 = 24 $ and $ f(3 + h)=3(3 + h)^2-(3 + h)=3(9 + 6h+h^2)-3 - h=27+18h + 3h^2-3 - h=24 + 17h+3h^2 $.
## Step 2: Substitute into the limit formula
$\lim_{h
ightarrow0}\frac{f(3 + h)-f(3)}{h}=\lim_{h
ightarrow0}\frac{(24 + 17h+3h^2)-24}{h}=\lim_{h
ightarrow0}\frac{17h + 3h^2}{h}=\lim_{h
ightarrow0}(17 + 3h)$.
## Step 3: Evaluate the limit
As $ h
ightarrow0 $, $ 17+3h
ightarrow17 $. So the slope of the tangent line is $ 17 $.
## Step 4: Find the equation of the tangent line
Using the point - slope form with the point $ (3,f(3))=(3,24) $ and $ m = 17 $, we have $ y - 24=17(x - 3) $, which simplifies to $ y=17x-51 + 24=17x - 27 $.
# Answer: The equation of the tangent line is $ y = 17x-27 $

### Problem 2
# Explanation:
## Step 1: Find $ f(2) $
Since the point $ (2,1) $ is on the graph of $ y = f(x) $, then $ f(2)=1 $.
## Step 2: Calculate $ f^{\prime}(2) $
The slope of the tangent line at $ x = 2 $ (which is $ f^{\prime}(2) $) can be found using the two points $ (2,1) $ and $ (5,13) $ on the tangent line. The slope formula is $ m=\frac{y_2 - y_1}{x_2 - x_1} $, so $ f^{\prime}(2)=\frac{13 - 1}{5 - 2}=\frac{12}{3}=4 $.
# Answer: $ f(2)=1 $ and $ f^{\prime}(2)=4 $

### Problem 3(a)
# Explanation:
## Step 1: Find the time when the object hits the ground
Set $ f(t)=0 $, so $ 256-16t^2 = 0 $. Then $ 16t^2=256 $, $ t^2 = 16 $, and $ t = 4 $ (we take $ t>0 $).
## Step 2: Use the average velocity formula
The average velocity $ v_{avg}=\frac{f(t_2)-f(t_1)}{t_2 - t_1} $. Here, $ t_1 = 0 $, $ t_2 = 4 $, $ f(0)=256-16(0)^2 = 256 $, $ f(4)=256-16(4)^2=256 - 256 = 0 $. So $ v_{avg}=\frac{0 - 256}{4 - 0}=\frac{-256}{4}=-64 $ ft/s.
# Answer: The average velocity is $-64$ ft/s

### Problem 3(b)
# Explanation:
## Step 1: Recall the definition of instantaneous velocity
The instantaneous velocity at time $ t $ is $ v(t)=\lim_{h
ightarrow0}\frac{f(t + h)-f(t)}{h} $. We want to find the velocity when $ t = 4 $ (when it hits the ground).
First, $ f(t)=256-16t^2 $, $ f(4 + h)=256-16(4 + h)^2=256-16(16 + 8h+h^2)=256-256-128h-16h^2=-128h-16h^2 $, and $ f(4)=0 $.
## Step 2: Substitute into the limit formula
$\lim_{h
ightarrow0}\frac{f(4 + h)-f(4)}{h}=\lim_{h
ightarrow0}\frac{-128h-16h^2-0}{h}=\lim_{h
ightarrow0}(-128 - 16h)$.
## Step 3: Evaluate the limit
As $ h
ightarrow0 $, $ -128-16h
ightarrow - 128 $. So the instantaneous velocity when it hits the ground is $-128$ ft/s.
# Answer: The instantaneous velocity when it hits the ground is $-128$ ft/s

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QUESTION IMAGE

Question

Explanation:

Problem 1(a)

Step 1: Find \( f(1) \) and \( f(2) \)

Step 2: Calculate the slope of the secant line

Step 3: Use point - slope form to find the equation

Step 1: Recall the definition of the slope of the tangent line

Step 2: Substitute into the limit formula

Step 3: Evaluate the limit

Step 4: Find the equation of the tangent line

Step 1: Find \( f(2) \)

Step 2: Calculate \( f^{\prime}(2) \)

Answer:

Problem 1(b)