Profit <- function(x){ (x-10)*(80-2*x) }Step 1. Find f(3): Step 2. Find f(-2): Step 3. Find f(x+c):
Sample.Function = function(x){8*x^3+7*x^2-5}
print(Sample.Function(3))## [1] 274print(Sample.Function(-2))## [1] -41Step 1: 274 Step 2: -41 Step 3 answer: f(x+c) = 8(x+c)^3 +7(x+c)^2 - 5
Step 1. Find lim as x approaches 1- f(x): Answer: 2
Step 2. Find lim as x approaches 1+ f(x): Answer: -5
Step 3. Find lim as x approaches 1 f(x): Answers: DOES NOT EXIST
library(Deriv)## Warning: package 'Deriv' was built under R version 3.3.2myf1=function(x){-2*x^3}
print(Deriv(myf1))## function (x)
## -(6 * x^2)myf1=function(x){(-8)/(x^2)}
print(Deriv(myf1))## function (x)
## 16/x^3myf1=function(x){5*x^(1/3)}
print(Deriv(myf1))## function (x)
## 1.66666666666667/x^0.666666666666667myf1=function(x){-2*x^(9/8)}
print(Deriv(myf1))## function (x)
## -(2.25 * x^0.125)Answer: -5/4 To find this answer, (35-40)/(4-0)
#As long as x does NOT equal zero.
my.avg=function(x){(630+2.4*x)/(x)}
print(my.avg)## function(x){(630+2.4*x)/(x)}my.derivative = function(x){(-2*x^(-2)+1)*(-5*x+9)}
print(Deriv(my.derivative))## function (x)
## 4 * ((9 - 5 * x)/x^3) - 5 * (1 - 2/x^2)Using the Quotient Rule = f’(x) = [((1/2)/(x(1/2))(-x3+1)-(-3x2)(5x(1/2) +7)]/(-x3+1)2 = (25x3+42x(5/2)+5)/(2x^(1/2)(x3-1)2)
my.derivative = function(x){(5*x^(1/2)+7)/(-x^3+1)}
print(Deriv(my.derivative))## function (x)
## {
## .e1 <- 1 - x^3
## .e2 <- sqrt(x)
## (2.5/.e2 + 3 * (x^2 * (5 * .e2 + 7)/.e1))/.e1
## }Using the Chain Rule: f’(x) = (4/3)[(3x^(-3) - 8x + 6)^(1/3)](-9*x^(-4) - 8)
fct1 <- function(x){(3*x^(-3) - 8*x + 6)^(4/3)}
DerivFct <- Deriv(fct1)
print(DerivFct)## function (x)
## -(1.33333333333333 * ((3/x^3 + 6 - 8 * x)^0.333333333333333 *
## (8 + 9/x^4)))Take the derivative of f(t).
fct1 <- function(x){(550*x^2)/( (x^2 + 15)^(0.5) )}
DerivFct <- Deriv(fct1)
print(DerivFct)## function (x)
## {
## .e1 <- x^2
## .e2 <- 15 + .e1
## 550 * (x * (2 - .e1/.e2)/sqrt(.e2))
## }Using the quotient and chain rule, f’(x) = [(1100t) * (t^2 + 15)^(0.5)-(t(t^2 + 15)^(-0.5) (550t^2)]/(t^2 + 15) To get the instantaneous change where t = 3, plug in f’(3) to get the answer F’(3) = 547.308. Rounded to whole number = 547
my.derivative = function(t){(550*t^2)/( (t^2 + 15)^(0.5) )}
print(Deriv(my.derivative))## function (t)
## {
## .e1 <- t^2
## .e2 <- 15 + .e1
## 550 * (t * (2 - .e1/.e2)/sqrt(.e2))
## }Step 1. What was the attendance during the third week into the season? Round your answer to the nearest whole number.
N(3) = 1000(6+0.1*3)^(0.5) = 2510 (rounded to the closest whole number)
print(round(1000*(6 + 0.1*3)^(0.5)),0)## [1] 2510Step 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.
Nt <- function(x){1000 * (6 + 0.1*x)^(0.5)}
Nt.Derived <- Deriv(Nt)
print(round(Nt.Derived(5)),0)## [1] 20Step 1. Use implicit differentiation to find dy/dx.
3x^3 + 4y^3 = 77 Using Implicit Differentiation and Chain Rule 9x^2 + 12y^2 * y’= 0 → 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
Using differentiation, we will determine if the function is increasing or decreasing. Using the quotient rule, f’(x) = -11 / (x-8)^2
However, given that denominator is (x - 8)^2. Any number squared is a positive number. Hence, f’(x) will always be negative except for when x = 8 as the function is undefined when x = 8 (cannot divide by zero).
Fct1 <- function(x){(x + 3) / (x - 8)}
Der.Fct1 <- Deriv(Fct1)
print(Der.Fct1)## function (x)
## {
## .e1 <- x - 8
## (1 - (3 + x)/.e1)/.e1
## }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.
To determine this, differentiate function 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.
Fct1 <- function(x){14 + (367*x^2) / (x^2 + 100)}
Der.Fct1 <- Deriv(Fct1)
print(Der.Fct1)## function (x)
## {
## .e1 <- x^2
## .e2 <- 100 + .e1
## 367 * (x * (2 - 2 * (.e1/.e2))/.e2)
## }Step 2. Over time, what temperature is the pizza approaching.
Take the limit of the function as it approaches infinity. Limit of the function can be separated into lim 14 + lim ((367t^2) / (t^2 + 100) ). The limit as t approaches infinity for 14 is 14. The limit of ((367t^2) / (t^2+100)) can be broken down into ((367*t^2) / [t^2*(1 + 100/t^2)], which can be reduced down into 367/(1+100/t^2). As the limit as t approaches infinity for 367/(1+100/t^2), this equals 367. Therefore, the limit for F(t) = 14 + 367 = 381.
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) = -.0014. Again, indicating local maximum.
Maximum number of packages = F(23) = .41 million packages On November 23rd
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)
Again, like question 19. 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)
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.
Area = 18432 ft^2 = x^2z (given square base). If z is height, bottom = x^2, top = x^2, and sides 4xz Total Costs = C = 5x^2 + 34xz + 3x^2 = 8x^2 + 12xz Using the Area formula, C = 8x^2 + 12x(18432/x^2) = 8*x^2 + 221184/x C’(x) = 16x - (221184/x^2) Set C’(x) = 0 and solve for x. X = Only one answer with a real number and x = 24. Therefore, z = (18432/24^2) = 32. So the dimensions are 24x24x32
Area = A = 1056 square years = xy Exterior fencing = $14.40/year and Interior fencing = $12.00/year Costs = C = 2x * $14.40 + 2y * $14.40 + 2x * $12.00 Y = 1056/x C = 52.8x + 30412.8/x C’ = 52.8 – 30412.8/x^2 Set C’ = 0 to find a local extrema. X turns out to equal + and – 24. But because we can’t have negative feet for fencing, only 24 is the answer. Thus x = 24. Plug in 24 for X into the Area formula and Y = 44.
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) = $31233.54
My assumption for that question was that the price depreciation was NOT going at an exponential rate (i.e., lost $1000 first year, then $4000, then $9000, etc. as then the following years, the price of the machine would be a negative value, which doesn’t make much sense, as the machines are worth at the very least $0).
Revenue = R = amount(x) * demand(p) Revenue = R(x) = x(23.2-0.4x) = 23.2x – 0.4x^2 To find the maximum value, take the derivative of R(x) → R’(x) = 23.2 – 0.8x Set R’(x) = 0. Solve for x. x = 29. 29 is the local maximum and is when the revenue is maximized.
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
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 Integrate the Marginal Profit Function to get the profit function. P(x) = 380x – 2x^2 + c. 1700 = 380(38) – 2(38)^2 + c. c = -9852 Hence, P(x) = 380x – 2*x^2 - 9852
Step 2. What is the profit from the sale of 56 clock radios? Profit(56) = 380 * (56) – 2*(56)^2 – 9852 = $5156
∫▒├ (-5(lna(y) )^3)/y┤dy
Set u = ln (y). Hence, du = 1/y dy. After substituting, we have Integral of -5 u^3 du. The antiderivative is (-5/4)(ln u)^4 + C. Replace U. The integral = (-5/4)(ln y)^4 + c
Take the definite integral as t goes from 0 to 9. The anti-derivative of P’(t) = 75t -6t^(1.5) + c. However, to find the increase, P(9)-P(0) which equals 513 wild animals.
Enter your answer below.
We are looking for the area between the two functions. 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. If we choose points x between 0 and 1, we note that the y value is higher in 6x^(1/2) than in 6x^2. Therefore, to find the bound region, we must take the area under 6*x^(1/2) and subtract the area under 6x^2 from it. To find the areas, we need to take the definite integrals.
The definite integral from 0 to 1 of 6*x^(1/2) is 4x^(3/2), or after plugging in the values, 4. The definite integral from 0 to 1 of 6x^2 is 2x^3, or after plugging in the values, 2. 4 – 2 = 2 (which is the answer to the question).
Using Separable differential equations, (1/y) dy = x^3 dx Integrate both sides → ln y = (x4)/4 eln(y) = y = e1 → y = e2 + C→ which is the solution to the this differential equation.
∫_(-7)^2▒├ x√( &x + 7) dx┤
Write your answer as a fraction. Via the Integration by Parts: [(10x(x+7)^(3/2) – 4*(x+7)^(5/2)]/15 From -7 to 2, the definite integral = -144/5
Step 1. How far has it traveled vertically at the moment when it hits the ground for the 20^(th) time?
The geometric series for a ball that bounces down then up: The geometric series: 2 * [sum as k approaches 20 is 92*(.22)^k ]– 46 (as the 1st drop goes down and does not return up a total of 46 meters). The first part is multiplied by two as it travels down and then back up, so we must keep in mind that this is part of the total distance. If we calculate this out → 2 * 46 [(1 - .22^(20))/(1 - .22)] – 46 = 71. 95 meters
Step 2. If we assume it bounces indefinitely, what is the total vertical distance traveled?
The geometric series: -46 + sum as k approaches infinity 92(.22)^k. General formula for Sum as k approaches infinity ar^k = (a) / (1-r)
Therefore, -46 + (92 / (1-.22)) = 71.95 meters
To the fifth degree, use the Taylor Polynomial x = 4. 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.
A=matrix(c(1,2,-1,2,1,1,-3,-3,2),3,3)
print(A)## [,1] [,2] [,3]
## [1,] 1 2 -3
## [2,] 2 1 -3
## [3,] -1 1 2Inverse.matrix = solve(A)
print(Inverse.matrix)## [,1] [,2] [,3]
## [1,] -0.8333333 1.1666667 0.5
## [2,] 0.1666667 0.1666667 0.5
## [3,] -0.5000000 0.5000000 0.5A=matrix(c(1,2,-1,2,1,1,-3,-3,2),3,3)
B= c(5,13,-8)
x <- solve(A)%*%B
print(x)## [,1]
## [1,] 7
## [2,] -1
## [3,] 0q=matrix(c(3,1,4,4,3,3,2,3,2),nrow=3)
print(q)## [,1] [,2] [,3]
## [1,] 3 4 2
## [2,] 1 3 3
## [3,] 4 3 2determinant.1 <- det(q)
print(paste("D = ",determinant.1))## [1] "D = 13"qx=matrix(c(1,4,5,4,3,3,2,3,2),nrow=3)
determinant.2 <- det(qx)
print(paste("Dx = ",determinant.2))## [1] "Dx = 19"qy=matrix(c(3,1,4,1,4,5,2,3,2),nrow=3)
determinant.3 <- det(qy)
print(paste("Dy = ",determinant.3))## [1] "Dy = -33"qz=matrix(c(3,1,4,4,3,3,1,4,5),nrow=3)
determinant.4 <- det(qz)
print(paste("Dz = ",determinant.4))## [1] "Dz = 44"The determinants are: 13 19 -33 44
q=matrix(c(3,1,4,4,3,3,2,3,2),nrow=3)
b=c(1,4,5)
x <- solve(q)%*%b
print(x)## [,1]
## [1,] 1.461538
## [2,] -2.538462
## [3,] 3.384615To solve the solution using R.