MATH 1425 Business Calculus

Jim Jack (J²)

MATH 1425 –Business Calculus

Grading

Attendance 1. Limits and the Derivative 1-1 Introduction to Limits

Cartesian Coordinate System

III IV

A function is a rule (process or method) that produces a correspondence between two sets of elements such that to each element in the first set, there is corresponds one and only one element in the second set. The first set is called the domain and the second set is called the range.

5 1 a w A x 10 2 b x B y 15 3 y C f x  x  2 x

f f(x)

 x  exactly one y  function (rule for finding y) For every input, there is exactly one output. f x  x  2

x g(x) -2 1 3 4 (6-1 Functions of two or more independent variables)

Heat Index – Temp and Humidity Wind Chill – Temp and Wind Speed GPA – Grades and Credit Hours Surfboard manufacturing Cx  500  70x Another board: Cy  200 100y Cx, y  700  70x 100y

z  f x, y  x  y

Cx, y  700  70x 100y f 10,5 

f x, y, z  2x2  3xy  3z 1 f 3,0,1  Revenue, Cost, Profit Suppose the surfboard company has demand equations given by: p  210  4x  y q  300  x 12y p is price of standard board, q is price of competition board, x and y are demand for the two boards. Find weekly revenue function R(x,y) and evaluate R20,10.

If cost is Cx, y  700  70x 100y, find profit. A company uses a box with a square base and open top for a bath assortment. If x is the length and y is the height, find the total amount of material required for a box M 5,10.

Cobb-Douglas production function f x, y  kxm yn ,m  n 1

A steel company has productivity at f x, y 10x0.2 y0.8, x is labor, y is capital. Find units of steel produced if the company uses 3000 units of labor and 1000 units of capital. Three-Dimensional Coordinate Systems Locate (–3, 5, 2) on the three-dimensional coordinate system. z

Consider the graph of z  x2  y2. If we let x  0, the equation becomes z  y2, which we know as the standard parabola in the yz plane. If we let y  0, the equation becomes z  x2, which we know as the standard parabola in the xz plane. The graph of this equation z  x2  y2 is a parabola rotated about the z axis. This surface is called a paraboloid.

Graphing Cross Sections Let f x, y  2x2  y2. Describe the cross sections in the planes y  0, y 1, y  2, y  3, and y  4, then in x  0, x 1, x  2, x  3, and x  4.

Limits f x  x  2

Limit lim f x  L f x  L x  c We write xc or   as if the functional value f x is close to the single real number L whenever x is close, but not equal, to c (on either side of c).

 x if x  0 f x    x if x  0

x f x    x

lim f x  x0 One-Sided Limits

lim f x  K We write xc and call K the limit from the left or the left-hand limit if f x is close to K whenever x is close to, but on the left of c on the real number line.

lim f x  L We write xc and call L the limit from the right or the right-hand limit if f x is close to K whenever x is close to, but on the right of c on the real number line. lim f x  L  Theorem 1 xc lim f x  lim f x  L xc xc For a two-sided limit to exist, the left-hand and right-hand limit must exist and be equal.

x x x lim  lim  lim  x0 x x0 x x0 x

lim f x  lim f x  lim f x  x1 x1 x1 lim f x  lim f x  lim f x  x2 x2 x2 Theorem 2 Properties of Limits: Let f and g be two functions such that: lim f x  L lim gx  M xc xc Where L and M are real numbers (both limits exist). Then lim k  k 1. xc for any constant k lim x  c 2. xc lim f x gx  lim f x lim gx 3. xc xc xc lim f x gx  lim f x lim gx 4. xc xc xc lim kf x  k lim f x 5. xc xc for any constant k 6. lim f x gx  lim f x  lim gx  LM xc xc  xc  lim f x f x   L 7. lim  xc  provided M  0 xc gx lim gx M xc n 8. lim n f x  n lim f x  L xc xc (the limit value must be positive for n even.) limx2  4x x3

Theorem 3 lim f x  f c for any f , a polynomial function xc nx nc lim  , dc  0. xc dx dc limx3  5x 1 x2

lim 2x2  3  x1

2x lim  x4 3x 1 x2 1 x  2 Let f x   x 1 x  2 lim f x  lim f x  lim f x  f 2  x2 x2 x2  

x2  4 lim  x2 x  2

x x 1 lim  x1 x 1

f x If lim f x  0 and lim gx  0, then lim xc xc xc gx is said to be indeterminate, or a 0 0 indeterminate form. Theorem 4 lim f x  L, L  0 lim gx  0 If xc and xc , then f x lim does not exist. xc gx

Limits of Difference Quotients

f x  h f x lim  h0 h f x  4x  5

f x  x  5

f x  x 1-2 Infinite Limits and Limits at Infinity 1 1 lim  lim  x1 x 1 x1 x 1

1 f( x )=  as x 1+ x -1

Since ¥ is not a real number, the limit above does not actually exist. We are using the symbol ¥ (infinity) to describe the manner in which the limit fails to exist, and we call this an infinite limit.

1 f( x )=� ギ as x 1- x -1

1 lim  x1 x 12 The vertical line x = a is a vertical asymptote for the graph of y = f (x) if f (x) ® ¥ or f (x) ® –¥ as x ® a+ or x ® a–. That is, f (x) either increases or decreases without bound as x approaches a from the right or from the left.

Locating Vertical Asymptotes Theorem 1 nx If f x  is a rational function such that dx dx  0, but nx  0, then the line x  c is a vertical asymptote of the graph of f. However, if both d(c) = 0 and n(c) = 0, there may or may not be a vertical asymptote at x = c. x2  x  2 Find vertical asymptotes for f x  . x2 1

x2  20 Find vertical asymptotes for f x  5x  22 Limits at Infinity The symbol ¥ can also be used to indicate that an independent variable is increasing or decreasing without bound.

If p is a positive real number, x p   as x  

Since the reciprocals of very large numbers are very small numbers, 1 1 lim  0. or  0 as x   x x p x p A line y = b is a horizontal asymptote for the graph of y = f (x) if f (x) approaches b as either x increases without bound or decreases without bound. Symbolically, y  b is a horizontal asymptote if lim f (x)  b or lim f (x)  b x x

Theorem 2 If p is a positive real number and k is any real number except 0, then k k lim  0 lim  0 x x p x x p lim kx p   lim kx p   x x Provided x p is a real number for negative x. 3 2 lim px Let px  2x  x  7x  3. Find x and lim px x

If n n1 p(x)  an x  an1x ⋯ a1x  a0 , an  0, n 1

n then lim px  lim an x   x x n and lim px  lim an x  , x x depending on an and n.

Describe the end behavior of: px  3x3  500x2 px  3x3  500x4 Finding Horizontal Asymptotes 3x2  5x  9 Consider the function f x  . 2x2  7

Theorem 3 If a xm  a xm1 ⋯ a x  a f (x)  m m1 1 0 , a  0, b  0 n n1 m n bn x  bn1x ⋯ b1x  b0 then a xm a xm lim f x  lim m and lim f x  lim m x x n x x n bn x bn x There are three possible cases for these limits.

1. If m < n, then the line y = 0 (x axis) is a horizontal asymptote for f (x). a 2. If m = n, then the line y  m is a horizontal bn asymptote for f (x) . 3. If m > n, f (x) does not have a horizontal asymptote. Find the horizontal asymptote: 5x3  2x2 1 f x  4x3  2x  7

3x4  x2 1 f x  8x6 10

2x5  x3 1 f x  6x3  2x2  7

2x2  5 Find all asymptotes of f x  x2  4x  4 1-3 Continuity A function f is continuous at a point x = c if lim f x ● xc exists ● f c exists, and lim f x  f c ● xc

A function f is continuous on the open interval (a,b) if it is continuous at each point on the interval. If one or more of these conditions are not met, the function is discontinuous. f x  x  2

x2  4 f x  x  2

x f x  x Continuous? f x  x  2 at x  2

x2  4 f x  at x  2 x  2

x f x  at x  0 and x 1 x If two functions are continuous on the same interval, then their sum, difference, product, and quotient are continuous on the same interval, except for values of x that make the denominator 0.

Theorem 1 Continuity Properties of some specific functions:

. A constant function is continuous for all x. n . For integer n > 0, f x  x is continuous for all x.

. A polynomial function is continuous for all x.

. A rational function is continuous for all x, except those values that make the denominator 0. n . For n an odd positive integer, f x is continuous wherever f (x) is continuous. n . For n an even positive integer, f x is continuous wherever f (x) is continuous and nonnegative. Continuous? f x  x2  2x 1

x f x  x  2x  3 f x  3 x2  4 f x  x  2

Solving Inequalities using Continuity

Theorem 2 If f is continuous on a,b and f x  0 for any x a,b, then either f x  0x a,b or f x  0x a,b.

x 1 Solve  0 x  2 1-4 The Derivative

Rate of Change The revenue (in dollars) from the sale of x planter boxes is given by Rx  20x  0.02x2,0  x 1,000

What is the change in revenue if production is changed from 100 to 400 planters?

What is the average change in revenue for this change? Average Rate of Change (Difference Quotient) For y = f (x), the average rate of change from x = a to x = a + h is f x  h f x f x  h f x m   ,h  0 x  h x h

A small steel ball dropped from a tower will fall a distance of y feet in x seconds, y 16x2.

Find the average velocity from x=2 to x=3 seconds.

Find and simplify the average velocity from x=2 to x  2  h seconds. h  0

Find the limit of the above as h  0, if it exists.

Interpret this limit. Slope of the graph Given y  f x, the slope of the graph at the f a  h f a point a, f a is given by lim , h0 h provided that limit exists. The slope of the graph is also the slope of the tangent line at a, f a.

The Derivative For y = f (x), we define the derivative of f at x, f x  h f x denoted f ¢(x), to be f x  lim h0 h if that limit exists. If f ¢(x) exists for each x in the open interval a,b, then f is differentiable over a,b. The derivative of the function f is a new function f ¢. The domain of f ¢ is a subset of the domain of f with the following interpretations: ■ For each x in the domain of f ¢, f ¢(x) is the slope of the line tangent to the graph of f at the point (x, f (x)). ■ For each x in the domain of f ¢, f ¢(x) is the instantaneous rate of change of y = f (x) with respect to x. ■ If f (x) is the position of a moving object at time x, then v = f ¢(x) is the velocity of the object at that time.

Given f x  4x  x2, find f ¢(x). Finding the Derivative To find f ¢ (x), we use a four-step process: Step 1. Find f (x + h)

Step 2. Find f (x + h) – f (x)

f x  h f x Step 3. Find h

f x  h f x Step 4. Find lim h0 h

Finding the tangent line slopes Given f x  4x  x2, find the slope of the graph at x  0, x  2, and x  3.

Graph y  f x  4x  x2, using the slopes above. Graphing calculator exercise Graph f and f ¢ (x) on the same axes. Is the graph of f ¢ (x) tangent to the graph of f x? Find the x intercept of graph of f ¢ (x). What is the slope of the graph of f for this value of x?

Given f x  x  2, find f ¢(x).

A company’s total sales (in $million) t months from now are given by St  t  2. Find and interpret S(25) and S ¢(25). Use this to estimate sales in months 26 and 27. Non-existence of the derivative The existence of the derivative at x  a depends on the existence of a limit at x  a. That is f a  h f a f a  lim must exist. If this h0 h limit does not exist, then f is non-differentiable at x  a, or f ¢(a) does not exist.

Let f x  x 1-5 Basic Differentiation Properties Theorem 1. f x  h f x The derivative f x  lim can h0 h dy be represented by f ¢ (x) or y ¢ or . dx

What is the slope of a constant function? If y  f x  C, then f ¢ (x) = 0.  f x  h  f x

Find the derivative of: f x  3 y  1.4 y   d 23  dx Power Rule A function of the form f x  xk is called a power function. This includes f (x) = x (where k = 1) and radical functions (fractional k).

Theorem 2. (Power Rule) Let y  f x  xn, where n is a real number, then y  f  x  dy  nxn1.   dx

Find the derivative of: f x  x5 y  x25 y  t 3

5 d x 3  dx 1 f x  x4 y  u

1 d  dx 3 x

Constant Multiple Property Let f x  kux , k is constant, u is differentiable at x. Step 1. f x  h  kux  h

Step 2. f x  h f x 

f x  h f x Step 3.  h

f x  h f x Step 4. lim  h0 h The derivative of a constant times a function is the constant times the derivative of the function.

Theorem 3 Let y = f (x) = k× u(x) be a constant k times a function u(x). Then y¢ = f ¢(x) = k × u ¢(x). Find the derivative of: f x  3x2

t3 y  6

1 y  2x4

0.4 d  dx x3 Sum and Difference Properties Let f x  ux vx ,ux and vx exist. Step 1. f x  h  ux  h vx  h

Step 2. f x  h f x 

f x  h f x Step 3.  h

f x  h f x Step 4. lim  h0 h

The derivative of the sum of two differentiable functions is the sum of the derivatives of the functions.

We can also show that: The derivative of the difference of two differentiable functions is the difference of the derivatives of the functions. Theorem 4. If y = f (x) = u(x) ± v(x), then y¢ = f ¢(x) = u¢(x) ± v¢(x).

Find the derivative of: f x  3x2  2x y  4  2x3  3x1 y  3 w  3w

 5 2 x3  d      dx 2 4  3x x 9  An object moves along the y axis so that its position at time x is f x  x3  6x2  9x.

Find the instantaneous velocity function vx.

Find the velocity at x  2 and x  5.

Find the time(s) when the velocity is zero. Let f x  x4  6x2 10. Find f ¢(x)

Find the equation of the tangent line at x = 1

Find the values of x where the tangent line is horizontal. 6-2 Introduction to Partial Derivatives

For a company producing only a standard surfboard: Cx  500  70x

The marginal cost is found by differentiating: Cx  70

For a company producing two types of surfboard: Cx, y  700  70x 100y Now suppose that we differentiate with respect to x, holding y fixed, and denote this by Cx x, y; or suppose we differentiate with respect to y, holding x fixed, and denote this by Cy x, y. Differentiating in this way, we obtain: Cx x, y  70, and Cy x, y 100.

Each of these is called a partial derivative, and in this example, each represents marginal cost. If z = f (x, y), then the partial derivative of f with respect to x is defined by  z f (x  h, y)  f (x, y)  lim  x h  0 h The partial derivative of f with respect to y is defined by  z f (x, y  k)  f (x, y)  lim  y k  0 k

Let z  f x, y  2x2  3x2 y  5y 1 Find: z x

fx 2,3 z y

f y 2,3 The profit function of the surfboard company was found to be: Px, y 140x  200y  4x2  2xy 12y2  700

Find Px 15,10 and Px 30,10. The productivity of a major computer manufacturer is approximated by the Cobb- Douglas production function f x, y 15x0.4 y0.6 x is labor, while y is capital. fx x, y is the marginal productivity of labor, while f y x, y is the marginal productivity of capital. Find these figures for 4,000 units of labor and 2,500 units of capital. 1-6 Differentials In section 1-4, we defined the derivative of f at x as the limit of the difference quotient: f x  h f x f x  lim h0 h Given y  x3, if x changes from 2 to 2.1, then y will change from 23  8 to 2.13  9.261. The change in x, called the increment in x, is denoted by x (run). Similarly, the change in y, called the increment in y, is denoted by y (rise). x  y 

Increments

x  x2  x1  x2  x1  x

y  y2  y1  f x2  f x1

 f x1  x f x1 x2 Given y  f x  2

Find x, y, and y / x for x1 1 and x2  2.

f x  x f x  Find 1 1 for x 1 and x  2. x 1

f x  x f x  y 1 1  x x y f x  lim x0 x Differentials y f x  y  f xx x

If y  f x is differentiable, then the differential is dy  f xdx or df  f xdx

Find dy for f x  x2  3x. Evaluate for: x  2 and dx  0.1 x  3 and dx  0.1 x 1 and dx  0.02 Since dy  f xdx it follows that y  dy.

Let y  f x  6x  x2 Find y and dy.

Graph y and dy for 1 x 1.

Compare y and dy for values in 1 x 1. A company manufactures and sells x transistor radios per week. If the weekly cost and revenue equations are: C(x)  5,000  2x x2 R(x) 10x  1,000 0  x  8,000 find the approximate changes in revenue and profit if production is increased from 2,000 to 2,010 units/week. 1- 7 Marginal Analysis in Business and Economics Marginal Cost, Revenue, Profit Definition: If x is the number of units of a product produced in some time interval, then Total cost = C(x) Marginal cost = C¢(x) Total revenue = R(x) Marginal revenue = R¢(x) Total profit = P(x) = R(x) – C(x) Marginal profit = P¢(x) = R¢(x) – C¢(x) = (marginal revenue) – (marginal cost) A company manufactures fuel tanks for cars. The total weekly cost of producing x tanks is: Cx 10,000  90x  0.05x2

Find the marginal cost function.

Find the marginal cost at a production level of 500 tanks per week.

Interpret above.

Find the exact cost of tank# 501.

Theorem 1 Cx  x Cx  Cx  1 1  Cx  x Cx  x 1 1 x 1 The research department of a company that manufactures MP3 players determined a price- demand equation as: x 10,000 1,000 p. The financial department provided a cost function: Cx  7,000  2x.

Find the domain of the price-demand equation.

Find and interpret the marginal cost function.

Find the revenue function as a function of x and determine its domain. Find the marginal revenue at x  2,000, x  5,000, and x  7,000, interpret.

Graph cost and revenue functions, find intersections and interpret.

Find profit function, its domain, and graph. Find the marginal profit at x 1,000, x  4,000, and x  6,000, interpret. Marginal Average Cost, Revenue, Profit If x is the number of units of a product produced in some time interval, then

Cost per unit: _ Cx Average cost per unit = Cx  x _ d Marginal average cost = Cx  Cx dx

Revenue per unit: _ Rx Average revenue per unit = Rx  x

_ d Marginal average revenue = Rx  Rx dx Profit per unit: _ Px Average profit per unit = Px  x

_ d Marginal average profit = Px  Px dx A small machine shop manufactures drill bits used in the petroleum industry. The manager estimates the daily cost to be: Cx 1,000  25x  0.1x2 _ _ Find Cx and Cx

_ _ Find C10 and C10, interpret.

Estimate average cost per bit at a production level of 11 bits per day.

Warning! To calculate the marginal averages you must calculate the average first (divide by x), and then find the derivative. If you change this order you will get no useful economic interpretations.