`sin^2x + cos^2x = 1`
`y = frac (-b +- sqrt(b^2 - 4ac)) (2a)`
`(x + 2)^2 + (y - 3)^2 = 16`
`slope = m = frac (\text(change in y)) (\text(change in x)) = frac (Deltay) (Deltax)`

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HCC Math212

(according to LWW)
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Ch 1.8 Functions & Models

What is Calculus?

  • Calculus is the study of change.
  • The study of how to quantify the change one variable experiences based on the change in another, related variable.
  • In mathematics, we express how two different variables are related using functions, for example ... `y=f(x)=3x^2 +1`
  • In mathematics, we quantify the relationship between the changes in two variables using fractions, for example ... `text(the change in y)/text(the change in x)`
  • The value of this fraction tells you how the change in `y` relates to the change in `x`

Where is Calculus Used?

Motion, electricity, heat, light, harmonics, acoustics, astronomy, dynamics, radioactive decay, birth and death rates, maximizing profits or revenue, minimizing cost, price elasticity of demand

Textbook Objectives for Section 1.8

  • Recognize and understand various business and economic terms
  • Create and evaluate mathematical models for real-life situations


LWW's Additional Objectives

  • Review necessary material on functions
  • Review necessary material on slope
  • Introduce necessary business terms: Cost, Revenue, Profit

Chapter 1 of this book is essentially all material that you are expected to know before taking this course. My experience with teaching this course is that it is time well spent for me to review with the students some of this material during the first two classes. I will focus on Section 1.8 mostly but will also touch on Section 1.6. I'm assuming you already know the following:

This is all in Chapter 1 of the text so if you need to review this material, you can find the information you need there.

Function Review

`f(x)` tells you that `f` is a function, and it takes `x` as an input. `f(x)` defines a relationship between two things that are represented mathematically by variables, `x` and `y`, where `y=f(x)`. I have seen many students in this course confused by the function notation, `f(x)`. I think it's just a result of not having much experience with functions, or maybe having a high school teacher who wasn't good at explaining what they are, or both. Anyway, when you see `f(x)` it is pronounced, "`f` of `x`". The `f` and the `x` are not separate; it's not `f` times `x` as some students want to think. They are one thing that defines a relationship between a variable `x` and a variable `y`. Where is `y`, you say? `y` comes from evaluating the function, `f(x)` at specific values of `x`. So `y` is `f(x)`; `y` is one thing and so is `f(x)`. Sometimes you will see just the `f` without the `(x)` part. Don't think that's something different; it's not. The `f` by itself refers to the right-hand side set of operations that are performed on the input to the function, whatever that input happens to be. A function has to have an input, and the input is whatever is inside the parentheses. Let's go through a couple of examples and hopefully, this will start to 'click' with you better.

Suppose you have a function, `f(x)=x^2 + 2x + 1`. This takes `x` as its input (what's inside the parentheses) and the right-hand side of the equation defines what the function does to the input, in this case, `x`. When you see this function, you should think:

  1. take the input to the function, `f`
  2. then square that input
  3. add 2 times the input to that result
  4. and finally add 1 to that result

Now consider the function, `g(x)=sqrt(4x+7)`. Again the input to this function, `g`, is just `x`. With this function you should think:

  1. take the input to the function, `g`
  2. multiply the input by 4
  3. then add 7 to that result
  4. and finally take the square root of that result


Remember to start inside parentheses or inside square roots and work you way out from there. In this course we will consider polynomial functions only of degree 3 or less, so constant, linear, quadratic, or cubic polynomial functions only. Later in the course we will also address exponential and logarithmic functions, but for now a quick review of linear, quadratic, and cubic polynomial functions is well worth the time.


Another confusing function thing for many people is when we have a function like, `f(x+1)`. In general, it is not `f(x) + 1`. Consider the function in the first example above, `f(x) = x^2 + 2x + 1`. Remember that whatever's inside the parentheses in the function notation defines what the input to the function is and the right-hand side of the equation defines what the function does to the input. It might be clearer to write the function this way,

`f(text(input))=text(input)^2 + 2*text(input) + 1`

Now it's clearer that regardless of what the input is, you know what to do with it. So

`f(x) = x^2 + 2x + 1`
`f(x+1) = (x+1)^2 + 2(x+1) + 1`
`f(text(giraffe)) = (text(giraffe))^2 + 2*(text(giraffe)) + 1`

One more thing, what about a function like `f(x) = 3`? This function says to take the input, whatever it is, do nothing with it and just set the function value, which is also the corresponding `y` coordinate, to the constant value of `3`. That's it; it's a constant function. Doesn't matter what the input to the function is, you always get `3` and only `3` as the output. It's graph is a horizontal line passing through `3` on the `y`-axis.


Linear Polynomial Functions

Consider the linear function, `f(x) = 5x + 3`. Here's what we know about this function just from the equation for `f(x)`.

  • It's a linear (1st degree polynomial) function.
  • Its graph is a line.
  • The slope of the line is 5 (a constant).
  • The `y`-intercept of the function is 3 (i.e., the point `(0,3)` ), found by evaluating `f(0)`.
  • The `x`-intercept of the function is `-3/5` (i.e., the point `(-3/5,0)`) [aka, a zero of `f(x)`; the solution to the equation `f(x) = 0`]
Quadratic Polynomial Functions

Consider the quadratic function, `f(x)=3x^2 - 6x - 9`. Here's what we know about this function from the equation for `f(x)`.

  • It's a quadratic (2nd degree polynomial) function.
  • Its graph is a parabola that opens up.
  • The slope of the parabola varies based on the value of `x`.
  • The `y`-intercept of the function is `-9` (i.e., the point `(0,-9)` ), found by evaluating `f(0)`.
  • The `x`-intercepts of the function are `3` and `-1` (i.e., the points `(3,0)` and `(-1,0)`) [aka, the zeroes of `f(x)`; the solution to the equation `f(x) = 0` solved by factoring, if possible, or using the quadratic formula]
Cubic Polynomial Functions

Consider the cubic function, `f(x)=x^3 - 3x^2 - 33x + 35`. Here's what we know about this function from the equation for `f(x)`.

  • It's a cubic (3rd degree polynomial) function.
  • Its graph is a curve with at most two direction changes (one less than its degree).
  • The slope of the function varies based on the value of `x`.
  • The `y`-intercept of the function is `35` (i.e., the point `(0,35)` ), found by evaluating `f(0)`.
  • The `x`-intercepts of the function are `-5`, `1`, and `7 ` (i.e., the points `(-5,0)`, `(1,0)`), and `(7,0)` [aka, the zeroes of `f(x)`; the solution to the equation `f(x) = 0`; solved in this course with a calculator]

"We must meet the challenge rather than wish it were not before us."

William Brennan, Jr.

Former Associate Justice of the US Supreme Court


This is where I wax philosophical and say that this is going to be a hard course, and it will require a lot of work from you to do well in it. Wishing the homework will just go away, or that you will somehow magically figure it out at the last moment, is a plan doomed to failure. Also, if your plan in this course is to memorize as much as you can and then regurgitate it out in a test, assume you will do very poorly in this course. That method doesn't work in calculus. The only way to truly do well in this course is to take things head on. Strive for understanding as much as you can (there is a reason for everything we will be doing, and I will be explaining ad nauseam these reasons as we go through it all) and strive to link as much as you can together and categorize. That's how you do well in this course. I tell my students each semester that if you look at each thing we work on and each problem we work, as a separate entity, you will at some point feel completely overwhelmed by the material. You have to look for and grasp the patterns, and when you see them group things together. Don't worry because I will be pointing out the similarities you need to recognize as we go. Some things that you group together will look completely different from the other when you take them on their own. Being able to categorize things as we go, we allow you to reduce this course material into something that is manageable; something that you can confidently deal with. That's how to approach this course.


Before I move on to the next section, let me just say that when you just read the short sections on linear, quadratic, and cubic polynomial functions, if you didn't try to come up with those same `x`- and `y`-intercepts that I gave you, then you need to go back and do it. Why? Because that's how you learn and know that you understand math.

"You need to get into the habit of working out for yourself things that you read in this course. Don't let a quiz or test be the first time you are actually trying to work out a problem with pencil and paper."

LWW (that's me)

Mathematical Modeling in Business and Economics

My experience teaching this course has been that a very large majority of the students taking it are business majors. After all, it is required for business majors. Because the students are weighted heavily to business, I have made as many of the examples I use and the problems that you will be asked to work business related rather than science related. The text we use for this course is geared in that same way; however, I go a little further with it than the text. In the text you can find problems asking you answer questions about the velocity or acceleration of a ball thrown off a cliff. Physics problems. I don't see the point to do this with business majors when I can find perfectly good business related problems that get at the same thing. So we will deal with things like cost, profit, and revenue functions; marginal cost and marginal revenue. We will maximize revenue and profit and minimize cost. To that end, I will now cover some business and economics terms that we will use throughout this course and some models that will be used frequently.

Term Symbol Description
Items (or units) `x` Number of items produced
Cost `C(x)` Total costs
Fixed costs `C(0)` Constant costs that do not depend on the number of items produced (rent, light, heat, etc.). (The `y`-intercept for `C(x)`)
Variable costs `C(x)-C(0)` Costs that depend on the number of items produced (labor, materials, etc.).
Revenue `R(x)` Income that depends on the number of items sold and the selling price per item.
Price `p` Selling price per item
Profit `P(x)` The difference between the revenue and the cost `[P(x)=R(x)-C(x)]`.
Break-even point   Point where revenue equals cost `[R(x)=C(x)`, or where `P(x)=0]`.
Supply function `p=S(x)` `p` is the price per item at which producers are willing to supply `x` items. (As supply of `x` increases, the price increases.)
Demand function `p=D(x)` `p` is the price per item at which consumers are willing to buy `x` items. (As demand of `x` increases, the price decreases.)
Equilibrium price `p_E` The price at which supply is equal to demand `[S(x)=D(x)]`.
Equilibrium point `(x_E,p_E)` The point at which the supply is equal to the demand. Consumers buy (demand) all `x_E` items supplied when the price is `p_E`.
Revenue in terms of the unit price
`{: (R(x),=,x*D(x)),( ,=, x*p) :}`
Revenue is the product of the number of items sold times the price per item. This price is set by the demand function, `p=D(x)`.

When you are dealing with a cost function, always assume that there is a fixed cost portion and a variable cost portion. Add the two together to get the total cost function. `C(0)` refers to the cost before any items are sold; that's why it's the fixed cost. It doesn't matter how many items are sold, you're paying that money. Cost, revenue, profit, supply, and demand functions will all be designated with capital letters. Price, `p`, and the numbers of items sold, `x`, are always lower-case. If the supply and demand functions seem weird, that's because they are written 'backwards' from what you might normally expect. You would probably normally think that supply and demand are functions of price, and they are. So you would have functions, `x = S(p)` and `x=D(p)`. We have solved both of those equations for `p` to make them compatible with the revenue, cost, and profit functions, all functions of `x`.


1.8.1 Modeling in Business

A soft drink company has fixed costs of $4,000 per day. The variable costs are $2.75 per case of soda. Each case sells for $5.25.

    • a. Write the cost function for producing `x` cases in a day

`C(x)` = Fixed costs per day + variable costs per day

`class{cmjx-highlight} {C(x) = 4000 + 2.75*x}`

    • b. Write the revenue function

`R(x)` = (number of items sold)(price per item)

`class{cmjx-highlight} {R(x) = x*5.25 = 5.25x}`

    • c. Write the profit function

`P(x)` = (Revenue for selling `x` items) - (Cost of selling `x` items)

`P(x) = R(x) - C(x) = 5.25x - (4000 + 2.75x)`

`class{cmjx-highlight} {P(x)=2.5x-4000}`


    • d. Find the break-even point

The break-even point occurs where revenue equals cost. We have the revenue and cost functions already so set them equal to each other and solve for `x`.

`R(x) = C(x) Rightarrow 5.25x = 4000 + 2.75x`

`2.5x = 4000 Rightarrow x = 4000/2.5 = 1,600`

So if the soft drink company makes `1,600` cases per day, they will break-even.

1.8.2 Equilibrium Point

Suppose that the supply function for a particular product is

`p=S(x)= x^2 + x`

and the demand function is


where `x` represents thousands of units and `p` represents thousands of dollars.


    • a. Find the equilibrium point `(x_E,p_E)`.

The equilibrium point occurs where supply equals demand, so ...

`{: (S(x),=,D(x), ), (x^2+x ,=,35-x, class{cmjx-note}{ text{substitute in for the functions}}), (x^2+2x-35,=,0, class{cmjx-note}{ text{gather like terms on left side}}), ( (x+7)(x-5),=,0, class{cmjx-note}{ text{factor the quadratic}}), (x,=,-7 or 5, class{cmjx-note}{ text{use zero-product rule}}) :}`

We have two possible answers, but a negative number makes no sense for the number of units sold so we throw out `-7` as a possible answer, and we have `x=5`. That is `5,000` units. Now we just need to find the corresponding price, `p_E`. We can find that by substituting the value we just found for `x_E` into either the `S(x)` or `D(x)` functions. If we use `D(x)`, because it's simpler, we get `p_E = 35 - 5 = 30`. That is `$30,000`. So,

`class{cmjx-highlight} {(x_E,p_E) = (5,30)}`
The equilibrium point occurs at `5,000` units when the price is `$30,000`.

1.8.3 Stock Market

Scott wants to buy a total of 500 shares inthe stock market. He will buy `x` shares at $4 per share and the rest at $6 per share.

    • a. Write the function for the total cost of the shares

We are told that `x` is the number of shares Scott buys at $4 per share. He wants to buy a total of 500 shares so the remaining shares will cost $6 per share. The number of these shares will be the difference of the total number bought, 500, and the number bought at $4 per share, `x`. So the number bought at $6 per share will be `500 - x`. We can now write the total cost to be (the number of $4 shares)(4) + (the number of $6 shares)(6) or:

`{: (C(x),=, 4x + 6(500-x)), (C(x) ,=, 4x+3000-6x), (C(x),=,3000-2x) :}`
`class{cmjx-highlight} {C(x)=3000-2x}`


    • b. Based on the model above, what will be Scott's cost if he buys 200 shares at $4?

We just need to substitute 200 in for `x`, which is the same as evaluating `C(x)` at `x=200`, and we get:

`C(200) = 3000 - 2(200) = 3000 - 400 = 2600`.

His total cost will be $2,600.

1.8.4 Phone Call Charges (piecewise functions)

Suppose that the cost of an overseas call is $9.00 for the first 3 minutes or less plus 95 cents for each additional minute.

    • a. Write a function for the cost of a call of x minutes. (Assume that a fraction of a minute over 3 minutes is charged the corresponding fraction of 95 cents.)

A major part of being able to work problems in this course is being able to take the problem stated in paragraph form, pick out what you are being asked to do, and translate that part into mathematics. Doing that sets you up for what your end goal is for the problem. If you can do that, you will know where you need to go and what you final answer looks like; you'll just need to figure out how to get there. Accomplishing this is a big deal because once you know what your goal is, then you can ask questions of yourself like, "What do I need to get there?" And then you can look for places in the rest of the problem statement for those answers.

Ok. So for this problem we have three sentences. The first sentence is just giving us some information; it's not asking us to do anything. The second sentence, however, starts with, "Write a function ..." That's what we're looking for; it's telling us what we need to do. So it says, "Write a function for the cost of a call ..." That tells us that we need to write a cost function! So now, the rest of the sentence says, "... for the cost of a call of `x` minutes." So we need to write a cost function, our variable is going to be `x`, which stands for the length of time of the call in minutes. So in more mathematical terms, we want ...

The cost funciton, `C(x)`, where `x` is the length of the call in minutes


The first and the third sentences are just giving us information. The first one starts, "Suppose that the cost ...". and the third one starts with, "Assume that a fraction ...". What's different about this problem from the first three we did is that two different functions are being described in the problem statement depending on the length of the phone call, and the deciding point between these two functions is 3 minutes. We have, "... the cost of an overseas call is $9.00 for the first 3 minutes or less ..." So that's describing a constant function. The cost is $9.00 if the call is 3 minutes or less. A call of length 1 second is $9.00, and a call of length 3.0 minutes is $9.00. So that's the first function, a constant function. We can write it as:

`C(x) = 9`, if `x <= 3`


Now we need to worry about what happens when `x>3`. We get that from the second half of the first sentence plus the third sentence. So we have, "... plus 95 cents for each additional minute (over 3)." Now the last sentence tells us to charge the additional minutes over 3 by the fraction of a minute at the rate of 95 cents per minute. That's important because that's different than being charged an additional 95 cents for any call with a length between 3 to 4 minutes, and another 95 cents for any call between 4 to 5 minutes. We are going to get charged for each fraction of a minute over 3 minutes. So we can write that function as:

`C(x) = 9+ 0.95(x-3)`, if `x>3`


We need `x-3` here because we want to apply this charge only to the portion over 3 minutes. If our call is 4 minutes, then we subtract 3; and we apply this additional charge only to 1 minute. So we have two different functions depending on whether `x<=3` or `x>3`. We want to write this as one function, and we do this with piecewise function notation as follows:

`class{cmjx-highlight} {C(x) = { {: (9, if x<=3), (9+.95(x-3),if x>3) :} }`


So now we have one function, `C(x)`, defined by two functions, one a constant function and the other a linear function, with a condition on the variable `x` that tells you when to use each one.

1.8.5 Perimeter


Building and fence Building and fence

A rectangular lot is fenced on three sides. An adjacent building forms the fourth side.

    • a. If the total length of the fencing is 48 feet, write a function that represents the area of the rectangle.

We are being asked to write a function representing the area of a rectangle; the area of a rectangle is found by multiplying the length of the rectangle by its width. So if a rectangle has length, `l`, and width, `w`, then we can write the function that gives the area, `A`, as:



At this point, I'll tell you that the statement of this problem leaves out some stuff'Stuff' is a mathematical term for stuff.. What they mean to say is, "Write a function in the simplest form that you can," which means in this case, in terms of the least number of variables. Anytime you can rewrite a function and reduce the number of variables on which it depends, you should do it. In what I cover in this course, we are going to deal with functions only of 1 variable. The problem with this equation of the area is that it has 2 variables in it so `A` is a function of 2 variables, `l` and `w`. So we need to rewrite `A` by eliminating either `l` or `w`. To do that you have to come up with something equivalent to either `l` or `w` that you can substitute into the area equation, and whatever you come up with should have the other variable in it. In other words, you need either to rewrite `l` in terms of `w`, or rewrite `w` in terms of `l`. So now think ... what haven't we used yet in this problem? The length of fencing! We are told in the statement of the problem that the total length of fencing used is 48 feet. So we should be looking to use that information to rewrite either `l` or `w`.

Two of the main characteristics of a closed shape are its area and its perimeter. If you place a flat shape on a table, then its area is the amount of space the shape covers, and its perimeter is the total distance around the edges of the shape. We can write an equation for the total length of fencing. From the picture there are 2 sides that have length `l` and only 1 side of length `w` because the opposite side of the enclosed area is the side of the building. So we can write the equation for the total length of fencing as:

`2l + w = 48`
This equation is referred to as a constraint equation for our original area equation. It is constraining what values `l` and `w` can take. For example, since the total length of fencing is 48 feet, the width, `w`, of the area has to be less than 48, and the length, `l`, has to be less than 24.


Now we have 2 equations and 2 unknowns (or variables) so we can use the second equation to eliminate one of the variables in the first equation.

`{: (2l + w,=,48), (w,=,48-2l) :}`
Now we can substitute `48-2l` in for `w` in the area equation.
`{: (A,=,l*w), (A,=,l*(48-2l)), (A,=,48l-2l^2) :}`
`class{cmjx-highlight} {A=48l-2l^2}`


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