Chapter 1.4 Functions Given by Simple Formulas

Not all functions can be given by a formula. In fact, many real world phenomenal cannot be expressed that way.

For now, we will focus on functions that can be expressed by a formula.

Formulas for Functions

If a function \(f\) is determined by an explicit formula, then it may be possible to evaluate the function \(f\) and determine the value \(f(x)\) for specific values of \(x\).

Example 1

Take \(f\) to be the function that is given by \[f(x) = \frac{2x+4}{(x-3)^2(x+1)}.\] Determine \(f(1)\), \(f(-2)\), and \(f(5)\).

We have

\[f(1)=\frac{2\cdot 1+4}{(1-3)^2(1+1)}=\frac{6}{8}=\frac{3}{4},\]

\[f(-2)=\frac{2\cdot (-2)+4}{(-2-3)^2(-2+1)}=\frac{0}{-25}=0,\]

\[f(5)=\frac{2\cdot 5+4}{(5-3)^2(5+1)}=\frac{14}{-24}=\frac{7}{12}.\]

Remember that evaluating a function at a finite number of points is only sampling the function. It can be useful for sketching the function, but it may be possible that you miss certain features of the function. For example, a function may be undefined at an \(x\) value we did not sample.

As a convention, when \(f\) is given by a formula but a domain for \(f\) is not specified, take the domain of \(f\) to be its maximal domain, the largest set on which the formula for \(f\) is meaningful.

Example 2

Take \(f\) to be the function that is given by \[f(x) = \frac{2x+4}{(x-3)^2(x+1)}.\] Determine \(\mathcal D(f)\), the domain of \(f\).

The formula for \(f\) involves adding multiplying and dividing numbers. Of those three operations, division by zero is the only one that won’t produce anything meaningful.

Therefore, \(\mathcal{D}(f)\) will contain all \(x\) for which the denominator is non-zero. The denominator of \(f\) is \((x-3)^2(x+1)\). Solve \[(x-3)^2(x+1)=0\] to get \(x=3\) or \(x=-1\).

Thus, \[\mathcal{D}(f)=\{x\in\mathbb{R}\colon x\not=3 \text{ and } x\not=1\}=(-\infty,-1)\cup(-1,3)\cup(3,\infty).\]

Simplifying formulas can be useful, but not when information is lost. Pay attention to the next example.

Example 3

Take \(f\) to be the function that is given by \[f(x) = \frac{x}{x}.\] What is the domain and range of \(f\)?

As long as \(x\not=0\), \(\frac{x}{x}=1\).

Therefore, \(f\) can be visualized like this:

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Thus, \[\mathcal{D}(f)=\{x\in\mathbb{R}\colon x\not=0 \}=(-\infty,0)\cup(0,\infty)\] and \(\mathcal{R}(f)=\{1\}.\)

Identifying properties of a function can be done through its formula. In this next example, we should that \(f\) is increasing.

Example 4

Take \(f\) to be the function with domain equal to \([0,3)\) that is given by \[f(x) = 2x+1.\] Show that \(f\) is increasing and use this property of \(f\) to determine the range of \(f\).

To show \(f\) is increasing, show that if \(x<y\), then \(f(x)<f(y)\). Equivalently, show that if \(x-y<0\), then \(f(x)-f(y)<0.\)

Calculate \(f(x)-f(y)\) to see that: \[\begin{align*} f(x)-f(y)&=[2x+1]-[2y+1]\\ &=2x-2y\\ &=2(x-y)\\ &<0. &&\text{ since }x-y<0 \end{align*}\]

Hence, \(f\) is increasing on \([0,3]\). Therefore, the range of \(f\) will be the open interval \([f(0),f(3)]\). Calculate \(f(0)\) and \(f(3)\) to obtain that \(f(0)=1\) and \(f(3)=7\). Thus the range is \([1,7]\).

The function in the previous example is a linear function. The coefficient of \(x\) plays an important role in whether the function is increasing or decreasing.

Linear Function and Slope

A linear function \(L\) is given by a formula of the form \[L(x) = mx+B,\] where \(m\) and \(B\) are real numbers.

The number \(m\) is the slope and the number \(B\) is the \(y\)-intercept.

A linear function has two terms: \(mx\) and \(B\). The \(mx\) tearm has an \(x\) which has a power of \(1\): \(x=x^1\).

There are other functions that have \(x\) terms, but with different powers.

Monic Monomial and Degree

A monic monomial is a function \(f\) of the form \[f(x) = x^n,\] where \(n\) is a natural number, or the function \[f(x) = 1.\]

In the first case, \(f\) is a degree \(n\) monomial, in the second, \(f\) is a degree 0 monomial

Understand the definition by completing the following example.

Example 5

Take \(L\) to be the linear function that is given by \[L(x) = 3x-5.\] Take \(f\) to be the monic polynomial that is given by \[f(x) = x^4.\] What is the slope and \(y\)-intercept of \(L\)? What is the degree of \(f\)?

The linear function \(L(x)=3x-5\) has slope \(m=3\) and \(y\)-intercept \(B=-5\). The degree of \(f(x)=x^4\) is \(n=4\).

In the next subsection we will focus specifically on linear functions and lines.

Lines

Here is an important fact about a line.

Two Points Determine a Line

Two points uniquely determine a line.

If a line \(L\) passes through the points \((a,b)\) and \((c,d)\), and \(a\ne c\), then \(L\) is a function.

We will derive the formula for a line using Euclidean Geometry.

Example 6

Assume that the line \(L\) passes through through the points \((a,b)\) and \((c,d)\), and \(a\ne c\). Find an equation that relates the variables \(x\) and \(y\) for any point \((x,y)\) in \(L\).

Take \((x,y)\) to be any point on the line that is not \((a,b)\) nor \((c,d)\). Create two right triangles. One right triangle is formed by connecting the points \((a,b),(c,d)\) and \((c,b)\). The other triangle is created by connecting the points \((a,b)\), \((x,y)\) and \((x,b)\).

Because the two triangles are similar, the ratio of their leg lengths must be equal. So \(\frac{d-b}{c-a}\) must equal \(\frac{y-b}{x-a}\):

\[\begin{align*} \frac{d-b}{c-a}&=\frac{y-b}{x-a}\\ \frac{d-b}{c-a}\cdot(x-a)&=y-b\\ m(x-a)&=y-b &&\text{ let }m=\frac{d-b}{c-a} \end{align*}\]

Therefore the formula is \(y-b=m(x-a)\).

The above example can be summarized in this example.

Point-Slope Formula

The line, \(L\), that passes through \((a,b)\) and \((c,d)\) with \(a\ne c\) has slope \(m\), where \[m = \frac{d-b}{c-a}\] and for any \((x,y)\) in \(L\), \[y-b = m(x-a).\]

This is the point-slope formula for a line.

Practice using the point slope formula with the following example.

Example 7

A line \(L\) has slope 3 and passes through \((2,7)\). For any \((x,y)\) in \(L\), write \(y\) in terms of \(x\)

The line has slope \(m=3\) and passes through \((2,7)\) (meaning \(a=2\) and \(b=7\)). Use point slope formula: \[\begin{align*} m(x-a)&=y-b\\ 3(x-2)&=y-7\\ 3x-3\cdot 2&=y-7\\ 3x-6&=y-7\\ 3x+1&=y. \end{align*}\]

The line is thus \(L(x)=3x+1\). Check by making sure it passes through \((2,7)\).

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In the next example, the information given is different compared to the previous example. Additional work is needed to translate the given information so that formula for the line can be created.

Example 8

Find the equation for the line that passes through the points \((2,3)\) and \((-3, 7)\).

The line passes through the points \((2,3)\) and \((-3,7)\). This is enough information to calculate the slope:

\[m=\frac{7-3}{-3-2}=-\frac{4}{5}.\]

Using point-slope formula (\(m(x-a)=y-b\)) we get

\[\begin{align*} -\frac{4}{5}(x-2)&=y-3\\ -\frac{4}{5}x+\frac{8}{5}&=y-3\\ -\frac{4}{5}x+\frac{23}{5}&=y.\\ \end{align*}\]

Thus the equation is \(y=-\frac{4}{5}x+\frac{23}{5}\).

The formula to get the equation of the line that passes \((a,b)\) and \((c,d)\) only works so long as \(a\ne c\). Do you see why?

Example 9

Suppose that \((a,b)\) and \((a,d)\) both lie on the line \(L\) and the points are distinct.

  1. Find an equation for \(L\).

  2. Is \(L\) a function?

The line passes through the points \((a,b)\) and \((a,d)\) is a vertical line. So the equation for \(L\) is \(x=a\). This is not a function.

Lines of the form \(x=a\) are called vertical lines.

We can rewrite the point/slope formula for a line \(L\) in a form that more easily shows where the line crosses the \(y\)-axis.

Slope y-intercept Form

This is the slope/\(y\)-intercept formula that relates \(x\) and \(y\) in an ordered pair \((x,y)\) in \(L\) by \[y = mx + B.\]

The number \(B\) is the \(y\)-intercept, it is the \(y\) coordinate of the intersection of \(L\) with the \(y\)-axis.

If the slope of \(L\) is not zero, then the \(x\)-intercept of \(L\) is the \(x\) value of the point where \(L\) intersects the \(x\)-axis.

Understand the above vocabulary but working through this next example.

Example 10

Find an equation for a line that passes through the points \((-1,5)\) and \((2, 14)\). Determine the line’s

  1. slope;

  2. \(y\)-intercept; and

  3. \(x\)-intercept.

  1. The slope of the line is \(m=\frac{14-5}{2-(-1)}=\frac{9}{3}=3.\)
  2. The \(y\)-intercept can be found by first writing an equation for \(L\): \[\begin{align*} 3(x+1)&=y-5\\ 3x+3&=y-5\\ 3x+8&=y. \end{align*}\] The \(y\)-intercept is \(8\).
  3. The \(x\)-intercept can be found by solving \(L=0\): \[\begin{align*} 3x+8&=0\\ 3x&=-8\\ x&=-\frac{8}{3}. \end{align*}\] The \(x\)-intercept is \(-\frac{8}{3}\).

In this section, we begin to build an elementary library of functions that we will use here and throughout the class.

An Elementary Library

To begin our elementary library, we define a monomial function.

Monomial Function and Degree

The function \(f\colon \mathbb R \to \mathbb R\) is said to be a monomial function if:

  • it is a constant function,

  • or if it is of the form \[f(x) = ax^n.\]

In this case, \(ax^n\) is said to be of degree \(n\).

These are very important functions, so we should know what they look like.

Example 11

For \(n\) equal to 1, 2, 3, 4, and 5, define by \(\mathrm{ pow}_n\) the function \[\mathrm{ pow}_n(x) = x^n.\] Sketch \(\mathrm{ pow}_n\) using a computer.

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The square root function, \(\mathrm{ pow}_{\frac{1}{2}}\), is important enough that we discuss it briefly now.

This function has the property that \[\mathrm{ pow}_{\frac{1}{2}}(x) = \sqrt{x} = y\] if and only if \[y^2 = x.\]

Try to sketch the function and determine its domain and range.

Example 12

Sketch \(\mathrm{ pow}_{\frac{1}{2}}\) and write down the domain and range of this function.

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The domain is \(D(f)=[0,\infty)\) and the range is \(R(f)=[0,\infty).\)

To end the section, see if you can do this example.

Example 13

Match \(\mathrm{ pow}_{\frac{1}{2}}(x)\), \(\mathrm{ pow}_1(x)\), \(\mathrm{ pow}_2(x)\), \(\mathrm{ pow}_3(x)\), \(\mathrm{ pow}_8(x)\), and \(\mathrm{ pow}_9(x)\) with the graphs sketched below.

Note that the sketches are not exact, but represent the shape of different functions accurately.

The graph VI is the graph of \(\textrm{pow}_{1/2}(x).\)

The graph II is the graph of \(\textrm{pow}_{1}(x).\)

The graph IV is the graph of \(\textrm{pow}_{2}(x).\)

The graph III is the graph of \(\textrm{pow}_{3}(x).\)

The graph V is the graph of \(\textrm{pow}_{8}(x).\)

The graph I is the graph of \(\textrm{pow}_{9}(x).\)

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