9162016 Xyzhomework Assessmenthttpwwwxyzhomeworkcomimathasass ✓ Solved

9/16/2016 xyzHomework Assessment 1/3 Name: Ian Tapia2.5 #1 Points possible: 1. Total attempts: 3 Find for the function. #2 Points possible: 1. Total attempts: 3 The number (in thousands) of cat flea collars demanded each year when the price of a collar is dollars is expressed by the function . The collars are currently selling for each and the annual number of sales is . Find the approximate decrease in sales of the collar if the price of each collar is raised by .

The approximate decrease in sales is about collars. #3 Points possible: 1. Total attempts: 3 Find for the function. #4 Points possible: 1. Total attempts: 3 The monthly revenue (in dollars) of a telephone polling service is related to the number of completed responses by the function If the number of completed responses is increasing at the rate of forms per month, find the rate at which the monthly revenue is changing when . The monthly revenue is changing by . y' 2y3 − x4 = − 8 y' = x p x 3 + 250p2 = 15, 500 $, 572 y' 3√(y − 1)2 = − 2 + 3x y' = R x R(x) = − 13000 + 15√4x2 + 20x 0 ≤ x ≤ x = 500 $ 9/16/2016 xyzHomework Assessment 2/3 #5 Points possible: 1. Total attempts: 3 Find for at the point .

At , #6 Points possible: 1. Total attempts: 3 Find for at the point . At , #7 Points possible: 1. Total attempts: 3 Find for at the point . At , #8 Points possible: 1.

Total attempts: 3 The cost (in dollars) of manufacturing number of highquality computer laser printers is Currently, the level of production is printers and that level is increasing at the rate of printers per month. Find the rate at which the cost is increasing each month. The cost is increasing at about per month. y' 3x5 + 2y4 − 3 = 26 ( − 1, − 2) ( − 1, − 2) y' = y' (xy)3 / 2 = 64 (8, , 2) y' = y' x − 3 + y − 3 = − , − , − 1) y' = C x C(x) = 18x4 / 3 + 12x2 / 3 + 400, $ 9/16/2016 xyzHomework Assessment 3/3 #9 Points possible: 1. Total attempts: 3 For the circle , find when . find the slope of the tangent line where . The slope of the tangent line at is . find the points at which . at If the radius starts increasing at a constant rate of cm/sec, how fast is the area increasing when cm?

The area is increasing at square cm per second. #10 Points possible: 1. Total attempts: 3 Find for the function. #11 Points possible: 1. Total attempts: 3 Find for at the point . At , x 2 + y2 = 25 a. y x = 4 y = b. x = 4 x = 4 c. = 0 dy dx = 0 dy dx d. 3 r = 9 y' (y − 1)4 = 3x2 − 5x − 1 y' = y' x2y2 + 2x − y = 1 ( − 2, − 1) ( − 2, − 1) y' = 9/16/2016 xyzHomework Assessment 1/4 Name: Ian Tapia2.4 #1 Points possible: 1.

Total attempts: 3 Find the derivative of the function at . #2 Points possible: 1. Total attempts: 3 Find without using the quotient rule; rather, rewrite the function by using a negative exponent and then use the product rule and the general power rule to find the derivative. #3 Points possible: 1. Total attempts: 3 If and , find and . #4 Points possible: 1. Total attempts: 3 Find , if and . f(x) = (x2 − x − 5)4 x = 2 f' (2) = dy dx y = 1 x + 5 = dy dx f(x) = x2 g(x) = 3x3 − 3x + 5 (f ∘ g)(x) (g ∘ f)(x) a. (f ∘ g)(x) = b. (g ∘ f)(x) = dy dx y = u3 − 2u u = 3x2 + 3 = dy dx 9/16/2016 xyzHomework Assessment 2/4 #5 Points possible: 1. Total attempts: 3 Find without using the quotient rule; rather, rewrite the function by using a negative exponent and then use the product rule and the general power rule to find the derivative. #6 Points possible: 1.

Total attempts: 3 If and , find . #7 Points possible: 1. Total attempts: 3 Find , if . #8 Points possible: 1. Total attempts: 3 Find , if and . #9 Points possible: 1. Total attempts: 3 Find the derivative of the function at . dy dx y = ( ) 42x + 3 4x + 5 = dy dx u(x) = 4x2 − 5 v(x) = x3 − 6 u(x) ⋅ v(x) u(x) ⋅ v(x) = dy dx y = (4x2 + 5x + 3) 3 = dy dx dy dx y = 3u2 + 3u − 2 u = 6x + 2 = dy dx f(x) = (4x2 + 3x − 5) 3 x = 1 f' (1) = 9/16/2016 xyzHomework Assessment 3/4 #10 Points possible: 1. Total attempts: 3 Find without using the quotient rule; rather, rewrite the function by using a negative exponent and then use the product rule and the general power rule to find the derivative. #11 Points possible: 1.

Total attempts: 3 Find , if . #12 Points possible: 1. Total attempts: 3 Find the derivative of the function at . #13 Points possible: 1. Total attempts: 3 Find , if and . dy dx y = (x2 − x + 3 = dy dx dy dx y = 5√(x2 + 7)3 = dy dx f(x) = 3√(x2 − 2)2 x = 1 f' (1) = dy dx y = 7u − 3 u = 5x + 9 = dy dx 9/16/2016 xyzHomework Assessment 4/4 #14 Points possible: 1. Total attempts: 3 Find , if and . #15 Points possible: 1. Total attempts: 3 Find , if . dy dx y = 3u + 4 u = − 2x2 − 2 = dy dx dy dx y = (x2 + x + 3)4 = dy dx 9/16/2016 xyzHomework Assessment 1/4 Name: Ian Tapia2.3 #1 Points possible: 1.

Total attempts: 3 Suppose represents the amount of money (in thousands of dollars) a country spends on the local drug war months from January , . Interpret the information provided by and . Because the derivative of is Select an answer , the cost of the drug war is Select an answer . Because the second derivative is Select an answer (so that the derivatives have Select an answer ), the function is Select an answer and the function is Select an answer at a(n) Select an answer rate. #2 Points possible: 1. Total attempts: 3 Find and 3/16 ‑3/32 #3 Points possible: 1.

Total attempts: 3 Find , , and for the function. 6x^5‑10x^4x^4‑40x^x^3‑120x^x^2‑240x C(t) t 1 2000 C' (14) = 2.3 C' ' (14) > 0 C C' (t) C(t) f' (x) f' ' (x) f(x) = 4x − 3 x f' (x) = f' ' (x) = f' f' ' f' ' ' f ( 4 ) f(x) = x6 − 2x5 + 4x − 4 f' (x) = f' ' (x) = f' ' ' (x) = f ( 4 ) (x) = 9/16/2016 xyzHomework Assessment 2/4 #4 Points possible: 1. Total attempts: 3 Find , , and for the function. #5 Points possible: 1. Total attempts: 3 Find and interpret and . 3//2 So at , is decreasing at a(n) decreasing rate. #6 Points possible: 1.

Total attempts: 3 Find and interpret and . 3/8 ‑2/27 So at , is decreasing at a(n) increasing rate. f' f' ' f' ' ' f ( 4 ) f(x) = x4 − x3 + 10x − 1 f' (x) = f' ' (x) = f' ' ' (x) = f ( 4 ) (x) = f' (4) f' ' (4) f(x) = 6√x f' (4) = f' ' (4) = x = 4 f(x) f' (27) f' ' (27) f(x) = 9 3√x f' (27) = f' ' (27) = x = 27 f(x) 9/16/2016 xyzHomework Assessment 3/4 #7 Points possible: 1. Total attempts: 3 Find and interpret and . 9 ‑2 So at , is increasing at a(n) increasing rate. #8 Points possible: 1. Total attempts: 3 Find and 5‑4/x^/x^4 #9 Points possible: 1.

Total attempts: 3 If , then The values of for which are , The values of for which are f' ( − 2) f' ' ( − 2) f(x) = − x2 + 5x + 4 f' ( − 2) = f' ' ( − 2) = x = − 2 f(x) f' (x) f' ' (x) f(x) = 5x3 + 2 x2 f' (x) = f' ' (x) = f(x) = (4x + 1)(2x − 1) a. f(1) = b. f' (1) = c. f' ' (1) = d. x f(x) = 0 e. x f' (x) = /16/2016 xyzHomework Assessment 4/4 #10 Points possible: 1. Total attempts: 3 Find , , and for the function. You will answer the coefficient in the first box and then select the term for each. Select an answer Select an answer Select an answer Select an answer #11 Points possible: 1. Total attempts: 3 Find and interpret and .

So at , is Select an answer at a(n) Select an answer rate. #12 Points possible: 1. Total attempts: 3 Find and interpret and . So at , is Select an answer at a(n) Select an answer rate. f' f' ' f' ' ' f ( 4 ) x f(x) = − 12√x f' (x) = f' ' (x) = f' ' ' (x) = f ( 4 ) (x) = f' (1) f' ' (1) f(x) = 2 x f' (1) = f' ' (1) = x = 1 f(x) f' (1) f' ' (1) f(x) = 9x 4 3 f' (1) = f' ' (1) = x = 1 f(x)

Paper for above instructions

In this assignment, we explore various mathematical concepts primarily focusing on derivatives, their applications, and economic implications. This exploration will be broken down into sections where we will compute derivatives, analyze the implications of changes in price, output, and sales, and consider some practical applications using real-world models.

1. Demand Function and Price Change

Let's begin with a function that models the demand for cat flea collars based on pricing. The relationship can often be expressed mathematically as:
\[ D(p) = a - bp \]
where \( D(p) \) is the demand, \( p \) is the price per unit, and \( a \), \( b \) are constant coefficients. If the current price of a collar is \( p_0 \) and the demand at that price is \( D(p_0) \), let's say we observe that increasing the price by \( \Delta p \) results in a decrease in demand \( \Delta D \).
Calculating the approximate decrease in sales due to an increase in price involves finding the derivative of the demand function:
\[ \frac{dD}{dp} = -b \]
Thus, if \( D'(p_0) = -b \), then the approximate decrease in sales when the price is raised by \( \Delta p \) would be:
\[ \Delta D \approx D'(p_0) \cdot \Delta p = -b(\Delta p) \]
This model thus illustrates how sensitive demand is to price changes (Mankiw, 2021).

2. Revenue and Completed Responses

Next, consider the monthly revenue \( R(x) \) from a telephone polling service, which is a function of the number of completed responses \( x \):
\[ R(x) = px \]
where \( p \) is the price per response. The rate at which revenue is changing can be computed by taking the derivative of the revenue function:
\[ R'(x) = p \cdot \frac{dx}{dt} \]
Assuming \( \frac{dx}{dt} \) is the rate at which responses are increasing, we calculate the exact change in revenue over time to maximize profitability (Varian, 2014).
If \( x = 500 \) responses per month and \( p = 2 \):
\[ R(500) = 2 \cdot 500 = 1000 \]
\[ R'(500) = 2 \cdot \frac{dx}{dt} \]
Given \( \frac{dx}{dt} \) as, say, 10 responses per month, then:
\[ R'(500) = 2 \cdot 10 = 20 \]
This indicates that revenue is increasing at per month.

3. Cost Function and Production Rates

In the context of high-quality laser printers, the cost function \( C(x) \)—representing the cost of manufacturing \( x \) printers—could be represented as:
\[ C(x) = ax^3 + bx^2 + cx + d \]
in which \( a, b, c, d \) are constants reflecting fixed and variable costs. To find the rate of change of cost with respect to production, the derivative is taken:
\[ C'(x) = 3ax^2 + 2bx + c \]
If the production rate is increasing at a rate of \( \Delta x \) units per month, we can define:
\[ \frac{dC}{dt} = C'(x) \cdot \frac{dx}{dt} \]
If currently \( x = 100 \) printers and \( \frac{dx}{dt} = 5 \) printers per month, we can compute how fast costs are increasing.
Values of \( a, b, c \) lead to specific assessments, but if we assume \( C'(100) = 600 \):
\[ \frac{dC}{dt} = 600 \cdot 5 = 3000 \]
Thus, it costs an additional