Identifying the Degree and Leading Coefficient of Polynomials

The formula just found is an example of a polynomial, which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as [latex]384 \pi ,[/latex] is known as a coefficient. Coefficients can be positive, negative, or zero and can be whole numbers, decimals, or fractions. Each product [latex]{a}_{i}{x}^{i},[/latex] such as [latex]384 \pi w,[/latex] is a term of a polynomial. If a term does not contain a variable, it is called a constant.

A polynomial containing only one term, such as [latex]5{x}^{4},[/latex] is called a monomial. A polynomial containing two terms, such as [latex]2x-9,[/latex] is called a binomial. A polynomial containing three terms, such as [latex]-3{x}^{2}+8x-7,[/latex] is called a trinomial.

We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient. When a polynomial is written so that the powers are descending, we say that it is in standard form.

Adding and Subtracting Polynomials

We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, [latex]5{x}^{2 },[/latex]and[latex]-2{x}^{2},[/latex]are like terms and can be added to get[latex]3{x}^{2},[/latex] but[latex]3x,[/latex]and[latex]3{x}^{2},[/latex]are not like terms and therefore cannot be added.

Multiplying Polynomials

Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials.

Difference of Squares

Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply [latex](x+1)(x-1),[/latex]using the FOIL method.

The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples.

Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term.

Performing Operations with Polynomials of Several Variables

We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example:

Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.

Using descriptive variables, we can notate these two functions. The function [latex]C(T)[/latex] gives the cost [latex]C[/latex] of heating a house for a given average daily temperature in [latex]T[/latex] degrees Celsius. The function [latex]T(d)[/latex] gives the average daily temperature on day [latex]d[/latex] of the year. For any given day, [latex]text{Cost}=C(T(d))[/latex] means that the cost depends on the temperature, which in turn depends on the day of the year. Thus, we can evaluate the cost function at the temperature [latex],T(d).[/latex] For example, we could evaluate [latex]T(5)[/latex] to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write [latex]C(T(5)).[/latex]

By combining these two relationships into one function, we have performed function composition, which is the focus of this section.

Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication, and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s income and then collecting all the data in a new column. If [latex]w(y)[/latex] is the wife’s income and [latex]h(y)[/latex] is the husband’s income in a year [latex]y[/latex] and we want [latex]T[/latex] to represent the total income, then we can define a new function.

If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write

Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.

For two functions [latex]f(x),[/latex] and [latex]g(x)[/latex] with real number outputs, we define new functions [latex]f+g,f-g,fg,[/latex] and [latex]\frac{f}{g}[/latex] by the relations

[latex](f + g)(x) = f(x) + g(x) [/latex]
[latex](f - g)(x) = f(x) - g(x) [/latex]
[latex](fg)(x) = f(x)g(x) [/latex]
[latex]\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \quad \text{where } g(x) \ne 0[/latex]

Create a Function by Composition of Functions

Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:

We read the left-hand side as “[latex]f[/latex] composed with [latex]g[/latex] at [latex]x[/latex]” and the right-hand side as “[latex]f[/latex] of [latex]g[/latex] of [latex]x.[/latex]” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol [latex]\circ ,[/latex] is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases [latex]f(g(x)) \ne f(x)g(x).[/latex]

It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first and then working to the outside. In the equation above, the function [latex]g[/latex] takes the input [latex]x[/latex] first and yields an output [latex]g(x).[/latex] Then the function[latex]f[/latex] takes [latex]g(x)[/latex] as an input and yields an output [latex]f(g(x)).[/latex]

In general,[latex]f \circ g[/latex] and [latex]g \circ f[/latex] are different functions. In other words, in many cases [latex]f(g(x)) \ne g(f(x))[/latex] for all [latex]x.[/latex] We will also see that sometimes two functions can be composed only in one specific order.

For example, if [latex]f(x)={x}^{2}[/latex] and [latex]g(x)=x+2,[/latex] then

[latex]f(g(x)) = f(x + 2) [/latex]
[latex]= (x + 2)^2 [/latex]
[latex]= x^2 + 4x + 4[/latex]

but

[latex]g(f(x)) = g(x^2) [/latex]

[latex]= x^2 + 2[/latex]

These expressions are not equal for all values of [latex]x[/latex] so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value [latex]x=\frac{1}{2}.[/latex]

Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.

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Evaluating Composite Functions

Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.

Evaluating Composite Functions Using Graphs

When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the [latex],xtext{-}[/latex] and [latex]ytext{-}[/latex] axes of the graphs.

How To

Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.

1. Locate the given input to the inner function on the [latex],xtext{-}[/latex] axis of its graph.
2. Read off the output of the inner function from the [latex],ytext{-}[/latex] axis of its graph.
3. Locate the inner function output on the [latex],xtext{-}[/latex] axis of the graph of the outer function.
4. Read the output of the outer function from the [latex],ytext{-}[/latex] axis of its graph. This is the output of the composite function.

Using a Graph to Evaluate a Composite Function

Using Figure 1, evaluate [latex],fleft(gleft(1right)right).[/latex]

Show Solution

To evaluate [latex],fleft(gleft(1right)right),,[/latex] we start with the inside evaluation. See Figure 2.

We evaluate [latex],gleft(1right),[/latex] using the graph of [latex],gleft(xright),,[/latex] finding the input of 1 on the [latex],xtext{-}[/latex] axis and finding the output value of the graph at that input. Here, [latex],gleft(1right)=3.[/latex] We use this value as the input to the function[latex],f.[/latex]

[latex]fleft(gleft(1right)right)=fleft(3right)[/latex]

We can then evaluate the composite function by looking at the graph of [latex],fleft(xright),,[/latex] finding the input of 3 on the [latex]xtext{-}[/latex] axis and reading the output value of the graph at this input. Here, [latex],fleft(3right)=6,,[/latex] so [latex],fleft(gleft(1right)right)=6.[/latex]

Analysis

Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value.

Try It

Using Figure 1, evaluate[latex],gleft(fleft(2right)right).[/latex]

Show Solution

[latex]gleft(fleft(2right)right)=gleft(5right)=3[/latex]

Finding the Domain of a Composite Function

As we discussed previously, the domain of a composite function such as [latex],fcirc g,[/latex] is dependent on the domain of [latex],g,[/latex] and the domain of [latex],f.[/latex] It is important to know when we can apply a composite function and when we cannot—that is, to know the domain of a function such as [latex],fcirc g.[/latex] Let us assume we know the domains of the functions [latex],f,[/latex]and[latex],g,[/latex] separately. If we write the composite function for an input [latex],x,[/latex] as [latex],fleft(gleft(xright)right),,[/latex] we can see right away that [latex],x,[/latex] must be a member of the domain of [latex],g,[/latex] in order for the expression to be meaningful because otherwise, we cannot complete the inner function evaluation. However, we also see that [latex],gleft(xright),[/latex] must be a member of the domain of [latex],f,[/latex]; otherwise, the second function evaluation in [latex],fleft(gleft(xright)right),[/latex] cannot be completed, and the expression is still undefined. Thus the domain of [latex],fcirc g,[/latex] consists of only those inputs in the domain of [latex],g,[/latex] that produce outputs from [latex],g,[/latex] belonging to the domain of [latex],f.[/latex] Note that the domain of[latex],f,[/latex] composed with [latex],g,[/latex] is the set of all [latex],x,[/latex] such that [latex],x,[/latex] is in the domain of [latex],g,[/latex] and [latex],gleft(xright),[/latex] is in the domain of [latex],f.[/latex]

Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.