1. It costs a toy retailer $10 to purchase a certain doll. He estimates that, if he charges x dollars per doll, he can sell 80-2x dolls per week. Find a function for his weekly profit.
Solution:
Profit per doll to sold = x * 10
Number of dolls sold per week = 80-2x
The function for his weekly profit will be:(x-10)(80-2x)
2. Given the following function: f(x)= 8x^3+ 7x^2-5
Solution:
f = function(x){8*x^3+7*x^2-5}
print(f(3))
## [1] 274
print(f(-2))
## [1] -41
find: f(x + c)
f(x+c) = 8(x+c)^3 +7(x+c)^2 - 5
f(x+c)=8x3+24cx2+24c2x+8c3+7x2+14cx+7c2-5
3.Use the graph to find the indicated limits. If there is no limit, state “Does not exist”.
Solution:
Find lim as x approaches 1- f(x): Answer: 2
Find lim as x approaches 1+ f(x): Answer: -5
Find lim as x approaches 1 f(x): Answers: Does not exists
4.Find the derivative for the following function.f (x)= -2x^3
Solution:
f=function(x){-6 * x^2}
print((f))
## function(x){-6 * x^2}
5.Find the derivative for the following function.f (x)=(-8)/x^2
Solution:
f=function(x){16 * x^3}
print((f))
## function(x){16 * x^3}
6.Find the derivative for the following function.g(x)=5x(1/3) Solution: 1.66666666666667/x0.666666666666667
7.Find the derivative for the following function.y= -2x^(9/8)
Solution:
f=function(x) {-(2.25 * x^0.125)}
print((f))
## function(x) {-(2.25 * x^0.125)}
8.Consider the graph of f(x). What is the average rate of change of f(x) from x1 = 0 to x2 = 4? Please write your answer as an integer or simplified fraction.
Solution:
m=(y2-y1)/(x2-x1)-> m=(35-40)/(4-0) m=-5/4
9.The cost of producing x baskets is given by C(x) = 630 + 2.4x. Determine the average cost function.As long as x does NOT equal zero.
Solution:
my.avg=function(x){(630+2.4*x)/(x)}
print(my.avg)
## function(x){(630+2.4*x)/(x)}
10.Use the Product Rule or Quote Rule to find the derivative. f(x) = (-2x(-2)+1)(-5x+9) Solution: Using the Product Rule f(x) = (4/(x3))(-5x+9) + (-5)(-2x(-2)+1) f(x)=(4(9-5x))/x^3-5(1-2/x2)
11.Use the Product Rule or Quotient Rule to find the derivative. f (x) = (5x^(1/2) + 7)/(-x^3 + 1)
Solution:
Using the Quotient Rule = f(x) = [((1/2)/(x(1/2))(-x3+1)-(-3x2)(5x(1/2) +7)]/(-x3+1)2 f(x)=(25x3+42x(5/2)+5)/(2x^(1/2)(x3-1)2)
12.Find the derivative for the given function. Write your answer using positive and negative exponents and fractional exponents.
Solution: f(x) = (3x^(-3) - 8x + 6)^(4/3). Using the Chain Rule: f(x) = (4/3)(3x^(-3) - 8x +6)^(1/3) f(x)=-(1.33333333333333((3/x^3 + 6 - 8 x)^0.333333333333333 *(8 + 9/x^4)))
13.After a sewage spill, the levelof pollution in Sootville is estimated by f(t) = (550t^2)/( (t^2 + 15)^(0.5) ), where t is the time in days since the spill occurred. How fast is the level changing after 3 days? Round to the nearest whole number.Take the derivative of f(t). Solution: (550x^2)/( (x^2 + 15)^(0.5))
f(3)=1010
Step 2. Determine N’(5) and interpret its meaning. Round your answer to the nearest whole number.
Using the Chain Rule, N’(t) = 500(6+0.1t)^(-0.5)*.1 = 50/(6+0.1t)^(0.5) N’(5) = 19.6. Rounded to the nearest whole number = 20 This is the rate of change of the number of people in attendance for week 5. This indicates that the rate of attendance is increasing at a rate of 20 compared to the last game. Solution:
n<- function(t){
return(round(50 / (6 + 0.1 * t)^(1/2),0))}
n(5)
## [1] 20
15.Consider the following function: 3x^3 + 4y^3 = 77
Solution: Step 1. Use implicit differentiation to find dy/dx.
y’ = -9x^2 / 12y^2 = -3x^2 / 4y^2
Step 2. Find the slope of the tangent line at (3, -1)
Slope at (3,-1) = -3(3)^2/(4*(-1)^2) = -6.75
16.Find the intervals on which the following function is increasing and on which it is decreasing. f(x) = (x + 3) / (x - 8)
Solution:
Using differentiation, we will determine if the function is increasing or decreasing. Using the quotient rule. f’(x) = -11 / (x-8)^2
17.A frozen pizza is placed in the oven at t = 0. The function f(t) = 14 + (367*t^2) / (t^2 + 100) approximates the temperature in degrees Fahrenheit of the pizza at time t. Solution: Step 1. Determine the interval for which the temperature is increasing and the interval for which it is decreasing. Please express your answers as open intervals.
F’(t)=Using the quotient rule, [(734* t) * (t^2 + 100) - (2t) (367*t^2)] / (t^2 + 100)^2, which simplifies into (73400*t) / (t2+100)2.
Given that t is time, time cannot be a negative number. And the denominator is a squared number, so NO number can negative as well. Therefore, from an interval of (0, infinity), the temperature is increasing.
function(x)=14 + (367*x^2) / (x^2 + 100)=381
18.A study says that the package flow in the East during the month of November follows f (x) = x^3/3340000 - (7x^2)/9475 + 42417727x/1265860000 + 1/33, where 1 £ x £ 30 is the day of the month and f (x) is in millions of packages. What is the maximum number of packages delivered in November? On which day are the most packages delivered? Round your final answer to the nearest hundredth.
Solution:
To find a local extrema, F’(x) = 0. F’(x) = (3x^2)/(3340000) - (14x)/(9475) + (42417727/1265860000). Solving for F’(x) = 0, there are two points. X = 23 and X = 1844249/1137. Because the question asks for x from day 1 to day 30, the only viable answer is X = 23. To check if this is a maximum, we can take the second derivative of the function, and if the 2nd derivative at x = 23 is negative (indicating concave downwards), this implies that this is a local extrema maximum. F’‘(x) = 6x/3340000 - 14/9475. F’’(23) = 0.41
Maximum number of packages = F(23) = .41 million packages On November 23rd
19.Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair. f (x) = 7x^2 + 28*x - 35 Solution: To find all local extrema, find the first derivative. F’(x) = 14x+28. Set F’(x) to equal 0 and solve for x. X = -2. To see if x at -2 is a local maximum or extrema, check the second derivative. If the second derivative at point x = -2 is positive, this is concave upwards and indicates local minimum and vice versa for the local maximum. F’‘(x) = 14. So F’’(-2) = 14, demonstrating a positive finding. Therefore, x is the local mimumum. Plugging in x = -2, y = -63. Answer: (-2,-63)
20.Use the Second Derivative Test to find all local extrema, if the test applies. Otherwise, use the First Derivative Test. Write any local extrema as an ordered pair. f (x) = -6x^3 + 27x^2 + 180x
Solution: F’(x) = -18x^2 + 54x + 180. Set F’(x) = 0 to find all extrema. X = -2 and 5 F’‘(x) = -36x + 54 Set F’‘(x) to equal zero to look for concave upwards or downwards. F’‘(-2) = 126 and F’’(5) = -126 Therefore, x at -2 is a local min and x at 5 is a local max. Plugging in -2 and 5. Answers (-2,-204) and (5,825)
21.A beauty supply store expects to sell 120 flat irons during the next year. It costs $1.60 to store one flat iron for one year. To reorder, there is a fixed cost of $6 , plus $4.50 for each flat iron ordered. In what lot size and how many times per year should an order be placed to minimize inventory costs?
Solution:
Let x = number of items per order (Yearly Storage cost per item) * (Average # of items Carried) = Yearly Carrying Cost ($1.60) * (x / 2) = 0.8 * x (Cost of each order) * (Number of orders placed per year) = Yearly Reordering Cost (6 + 4.50 * x) * (120 / x) = Yearly Reordering Cost Total Inventory Cost = (yearly carrying cost) + (yearly reordering cost) Y = (0.8 * x) + (720 / x) + 540 Take the derivative of Y to find a local extrema. Y’ = 0.8 - 720/(x^2). Set Y’ = 0. Solve for x. X = 30. Therefore, the answer is 4 sets of 30.
23.A farmer wants to build a rectangular pen and then divide it with two interior fences. The total area inside of the pen will be 1056 square yards. The exterior fencing costs $14.40 per yard and the interior fencing costs $12.00 per yard . Find the dimensions of the pen that will minimize the cost.
Solution:
Area=1056 square years Exterior fencing = $14.40/year Interior fencing = $12.00/year Costs = C = 2x * $14.40 + 2y * $14.40 + 2x * $12.00 x = 24 and Y = 44
24.It is determined that the value of a piece of machinery declines exponentially. A machine that was purchased 7 years ago for $67000 is worth $37000 today. What will be the value of the machine 9 years from now? Round your answer to the nearest cent.
Solution:
Price at time X = P(x), which declines its price at an exponential decay. Therefore P(0) = $67000. As noted above, P(7) = $37000 A general formula for P(x) = P0 * e^(kt), where decay occurs when k < 0. Plug in the numbers. 37000 = 67000 e ^(k7). Solve for k. k = -0.848 Now P(9) = 67000 e ^ (-0.848 * 9) = $17244.50
25.The demand function for a television is given by p = D(x) = 23.2 - 0.4x dollars. Find the level of production for which the revenue is maximized.
Solution: Revenue = R = amount(x) * demand(p) Revenue = R(x) = x(23.2-0.4x) = 23.2x - 0.4x^2 R’(x) = 23.2 - 0.8x Solve for x = 29
26.The amount of goods and services that costs $400 on January 1, 1995 costs $426.80 on January 1, 2006 . Estimate the cost of the same goods and services on January 1, 2017. Assume the cost is growing exponentially. Round your answer to the nearest cent.
Solution:
Use the exponential growth formula = P(t) = P0 * e^(kt). If you plug in for January 1, 2006 (which is t = 11 in years), $426.80 = $400 e ^ (11k). Solve for k (divide 400 on both sides, and then take the ln of both sides, then divide by 11), k = 0.00589. Plug in for t = 22. $400 * e ^ (0.00589 * 22) = $455.34
27.A manufacturer has determined that the marginal profit from the production and sale of x clock radios is approximately 380 - 4x dollars per clock radio.
Solution: Step 1. Find the profit function if the profit from the production and sale of 38 clock radios is $1700.
Marginal Profit Function = P’(x) = 380 - 4x P(x) = 380x - 2x^2 + c. 1700 = 380(38) - 2(38)^2 + c c = -9852
Step 2. What is the profit from the sale of 56 clock radios? Profit(56) = 380 * (56) - 2*(56)^2 - 9852 = $5156
28.Use integration by substitution to solve the integral below.
Solution:
Set u = ln (y). Hence, du = 1/y dy. substituting we have Integral of -5 u^3 du. The antiderivative is (-5/4)(ln u)^4 + C
29.It was discovered that after t years a certain population of wild animals will increase at a rate of P´(t) = 75 - 9t^(1/2) animals per year. Find the increase in the population during the first 9 years after the rate was discovered. Round your answer to the nearest whole animal.
Solution: The anti-derivative of P’(t) = 75t -6t^(1.5) + c
30.Find the area of the region bounded by the graphs of the given equations. y = 6x^2,y = 6???x
Solution: First, to find the intersection points, set 6x^2 = 6x^(1/2) and solve for x. X has two answers; 0 and 1. Therefore, the regions are bounded at x = 0 and x = 1. The definite integral from 0 to 1 of 6x^2 is 2x^3, or after plugging in the values, 2. 4 - 2 = 2
31.Solve the differential equation given below. dy/dx = x^3 y
Solution:
Using Separable differential equations, (1/y) dy = x^3 dx Integrate both sides ln y = e(x4/4)+ C
32.Use integration by parts to evaluate the definite integral below.
Solution: the Integration: [(10x(x+7)^(3/2) - 4*(x+7)^(5/2)]/15 From 2 to -7 = -144/5
34.Find the Taylor polynomial of degree 5 near x = 4 for the following function. y = 3e^(5x - 3)
Solution: Plug into the formula 3e^(17) + 15e^(17) * (x - 4) + (75/2)e^(17) (x - 4)^2 + (375/6)e^(17) (x - 4)^3 + (1875/24)e^(17) (x - 4)^4 + (9375/120)e^(17) (x - 4)^5.