-– title: “R Notebook” output: hw_1 —

Test Name: homework1(Test)

  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. f(x) = 80 - 2(10)
f <- function(x) (80-2*x)*(x-10)
f
## function(x) (80-2*x)*(x-10)
  1. Given the following function: f x = 8x3 + 7x2 - 5
f <- function(x) 8*x^3+7*x^2-5
#Step 1. Find  f 3. 
f(3)
## [1] 274
#Step 2. Find  f -2.  
f(-2)
## [1] -41

Step 3. Find f x + c. We can move it left or right by adding a constant to the x-value:

  1. Use the graph to find the indicated limits. If there is no limit, state “Does not exist”.
    Step 1. 2 Step 2. 5 Step 3. DNE

  2. Find the derivative for the following function. f x = -2x3

Deriv(~ (-2)*x^3, "x")
## -(6 * x^2)
  1. Find the derivative for the following function. f x = -8x2
Deriv(~ (-8)/(x^2), "x") 
## 16/x^3
  1. Find the derivative for the following function. gx = 53x
Deriv(~ (5)*x^(1/3), "x")
## 1.66666666666667/x^0.666666666666667
  1. Find the derivative for the following function. y = -2x98
Deriv(~ (-2) * x^(9/8), "x")
## -(2.25 * x^0.125)

  1. 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. (35-40)/(4-0) = -5/4

  2. The cost of producing x baskets is given by Cx = 630 + 2.4x. Determine the average cost function.

f <- function(x) (630+2.4*x)/(x)
f
## function(x) (630+2.4*x)/(x)
  1. Use the Product Rule or Quotient Rule to find the derivative.
    f x = -2x-2 + 1 -5x + 9
Deriv(~ (-2*x^(-2)+1)*(-5*x+9), "x")
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)
  1. Use the Product Rule or Quotient Rule to find the derivative.
    f x = 5x12 + 7-x3 + 1
Deriv(~ (5*x^(1/2)+7)/(-x^3+1), "x")
## {
##     .e1 <- 1 - x^3
##     .e2 <- sqrt(x)
##     (2.5/.e2 + 3 * (x^2 * (5 * .e2 + 7)/.e1))/.e1
## }
  1. Find the derivative for the given function.  Write your answer using positive and negative exponents and fractional exponents instead of radicals.
    f x = 3x-3 - 8x + 643
Deriv(~ (3*x^(-3)-8*x+6)^(4/3), "x")
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 * 
##     (8 + 9/x^4)))
  1. After a sewage spill, the level of pollution in Sootville is estimated by  f t = 550t2 t2 + 15, 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. 
f <- function(t) round((550*t^2)/(t^2+15)^(1/2))
fprime <- function(t) Deriv(~ (550*t^2)/(t^2+15)^(1/2), "t")
f(3)
## [1] 1010
  1. The average home attendance per week at a Class AA baseball park varied according to the formula Nt = 10006 + 0.1t12 where  t is the number of weeks into the season 0 £ t £ 14 and  N represents the number of people.

Step 1. What was the attendance during the third week into the season? Round your answer to the nearest whole number.

N <- function(t) round(1000 * (6 + 0.1 * t)^(1/2))
N(3)
## [1] 2510

Step 2. Determine N ´5 and interpret its meaning. Round your answer to the nearest whole number.

N1 <- function(t) round(50 / (6 + 0.1 * t)^(1/2))
N1(5)
## [1] 20
  1. Consider the following function:
    3x3 + 4y3 = 77
#Step 1. Use implicit differentiation to find dydx.  
#dy(x)/d(x) = 3*x^2/4*y^2
#Step 2. Find the slope of the tangent line at 3, -1.  
#dy(x)/d(x) = 3*(3)^2/4*(-1)^2 = 18/4
  1. Find the intervals on which the following function is increasing and on which it is decreasing.
    f x = x + 3x - 8
Deriv(~ (x+3)/(x-8), "x")
## {
##     .e1 <- x - 8
##     (1 - (3 + x)/.e1)/.e1
## }
#(-∞,8),(8, +∞)
  1. A frozen pizza is placed in the oven at t = 0.  The function Ft = 14 + 367t2t2 + 100  approximates the temperature (in degrees Fahrenheit) of the pizza at time t . 
#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. 
Deriv(~ 14+(367*t^2)/(t^2+100), "t")
## {
##     .e1 <- t^2
##     .e2 <- 100 + .e1
##     t * (734 - 734 * (.e1/.e2))/.e2
## }
Deriv(Deriv(~ 14+(367*t^2)/(t^2+100), "t"))
## {
##     .e1 <- t^2
##     .e2 <- 100 + .e1
##     .e3 <- .e1/.e2
##     .e4 <- .e2^2
##     .e9 <- 734 + 734 * (1 - .e3) - 734 * .e3
##     c(.e1 = -(t * .e9/.e4), t = (734 - .e1 * (2 * .e9 + 734)/.e2)/.e2, 
##         .e2 = t * (1468 * .e3 - 734)/.e4)
## }
#t is increasing from (0, +∞)

#Step 2. Over time, what temperature is the pizza approaching? 
f <- function(t) 14+(367*t^2)/(t^2+100)
#lim as t -> inf = 381 degrees F
  1. A study says that the package flow in the East during the month of November follows  f x = x33340000 - 7x29475 + 42417727x1265860000 + 133, 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. 
f <- function(x) (x^3/3340000 - 7*x^2/9475 + 42417727*x/1265860000 + 1/33)
#the equation is f'(x)
fprime <- function(x) Deriv(~ (x^3/3340000 - 7*x^2/9475 + 42417727*x/1265860000 + 1/33), "x")

fprime(0)
## 0.0335090191648366 + x * (8.98203592814371e-07 * x - 0.00147757255936675)
#solving for f'(x) = 0, we get 23 as a root
#0 = 0.0335090191648366 + x * (8.98203592814371e-07 * x - 0.00147757255936675) 

#then we plug it into the original equation to get the number of packages
f(23)
## [1] 0.4138353
#approximately 0.42 million packages 
  1. 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 = 7x2 + 28x - 35
f <-function(x) expression(7*x^2 + 28*x - 35)
fprime <- function(x) D(expression(7*x^2 + 28*x - 35), 'x')
fsecondprime <- function(x)D(D(expression(7*x^2 + 28*x - 35), 'x'), 'x')

# when f'(x) = 0, x = -2
f(-2)
## expression(7 * x^2 + 28 * x - 35)
#solving for f(-2) = (7 * x^2 + 28 * x - 35); we get 63

#local min(-2, 63)
  1. 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
f <-function(x) (-6*x^3 + 27*x^2 + 180*x)
fprime <-function(x) D(expression(-6*x^3 + 27*x^2 + 180*x), 'x')
fdoubleprime <-function(x) D(D(expression(-6*x^3 + 27*x^2 + 180*x), 'x'), 'x')

fprime
## function(x) D(expression(-6*x^3 + 27*x^2 + 180*x), 'x')
#when f'(0) = -6*x^3 + 27*x^2 + 180*x; x(-2, 5)

fdoubleprime(-2)
## 27 * 2 - 6 * (3 * (2 * x))
fdoubleprime(5)
## 27 * 2 - 6 * (3 * (2 * x))
#when f''(x) 54 - 36x we plug in x(5, -2) -> (-126, 126)

f(-2)
## [1] -204
f(5)
## [1] 825
#for the pair our local min is (-2, 204) and max (5, 825)
  1. 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? 
f<-function(x) (120/x*1.6+x*(6+4.5*120/x))

fprime <-function(x) Deriv(~ 120/x*1.6+x*(6+4.5*120/x), 'x')
fsecondprime <-function(x)Deriv(~ 6 - 192/x^2, 'x')

fprime(0)
## 6 - 192/x^2
#if f'(0) is 30, then we plug in to f(30)
f(30)
## [1] 726.4
#then we divide f(30)/120 to get 726.4/120 is approximately 6
#so a lot size of 30 6 times a year
  1. A shipping company must design a closed rectangular shipping crate with a square base.  The volume is 18432 ft3.  The material for the top and sides costs $3 per square foot and the material for the bottom costs $5 per square foot.  Find the dimensions of the crate that will minimize the total cost of material. 
# 18432 = (base = b^2, side = h, $3/ft^2)
#18432 = x^2y

#f<-function(x) (x^2y)
# (32, 24) 
  1. 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.
#1056 = x*y
#[exterior]14.40*(2y+2x)+ [interior]12*(2x) 
#plugging in y as 1054/x;
#14.40(2(1054/x)+2x)+12(2x)
#(24,44)
  1. 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.
#ab^t 400->426.80 in 11 years
f <- function(t) 67000*((37000/67000)^(t/7))
#7 + 9 = 16
f(16)
## [1] 17244.5
  1. The demand function for a television is given by p = Dx = 23.2 - 0.4x dollars.  Find the level of production for which the revenue is maximized. 
#-23.2/0.4 = 29
  1. 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.
#ab^t 400->426.80 in 11 years
f <- function(t) 400*(426.80/400)^(t/11)
f(11)
## [1] 426.8
#The future value below
f(22)
## [1] 455.3956
  1. 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.

Step 1. Find the profit function if the profit from the production and sale of 38 clock radios is $1700.

#Integral of 380-4x dx = 380x-2*x^2
#1700 = 380*38-2*(38^2)+c; c=-9852

Step 2. What is the profit from the sale of 56 clock radios?

#The profit for 56 is 380*56-2*(56)^2 - 9852
#The profit is $5156
  1. Use integration by substitution to solve the integral below.

-5ln(y)^3/y dy

#u = ln(y); du/dy = 1/y
#integral of -5ln(y)^3/y dy = -(5/4)ln(y)^4+C
  1. It was discovered that after t years a certain population of wild animals will increase at a rate of P´t = 75 - 9t12 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. 
#The integral from 0 to 9 is (75-9sqrt(t) dt) = 513
  1. Find the area of the region bounded by the graphs of the given equations.
    y = 6x2, y = 6x Enter your answer below.
#A = integral from 0 to 1 of 6x (upper) - 6x^2(lower) 
# integral of 6x^2 from 0 to 1 is 2 and integral of 6x is 3
# the area is 1
  1. Solve the differential equation given below.
    dydx = x3y
#x^3 -> x^4/4
#e^((x^4)/4) + c
  1. Use integration by parts to evaluate the definite integral below.
    Write your answer as a fraction.
#from -7 to 2 for x*sqrt(x+7)
#u=x, du=√x+7dx √dx, u=2/3(x+7)^2/3
#=-144/5
  1. The following can be answered by finding the sum of a finite or infinite geometric sequence.  Round the solution to 2 decimal places.
    A rubber ball is dropped from a height of 46 meters, and on each bounce it rebounds up 22 % of its previous height.

Step 1. How far has it traveled vertically at the moment when it hits the ground for the 20th time?

# from 1 to 20 for 46*(.22)^x
#comes out to 12.9743589....
#rounds out to 12.97

Step 2. If we assume it bounces indefinitely, what is the total vertical distance traveled?

#the ball has been displaced 46 meters and bounces
# c + 2(Sigma(1, ∞) for (46*(.22)^x)
# comes out to 46+2(12.97) = 71.94
  1. Find the Taylor polynomial of degree 5 near x = 4  for the following function.
    y = 3e^(5x - 3)
#this function incepted me back to freshman year of college.  It reminds me of a maclaurin series for pi
#It was a foggy morning, and as usual I was trying to memorize ln / e and trig shortcuts... 
#3*(e^17)+15*(e^17)*(x-4)+75/2*(e^17)*(x-4)^2+125/2*(e^17)*(x-4)^3+625/8*(e^17)*(x-4)^4+625/8*(e^17)*(x-4)^5