\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Precalculus - Functions

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The idea of a function is that it takes a value as an input and gives exactly one output value. The output value may change based on the input. The key idea here is the one and only value can be output by the the function. On a graph, this requirement is verified using the vertical line test. Many books and videos model this idea using a black box. However, most of the time, you will know what is going on with the function.

Function Notation

One of the nice consequences of requiring that there be only one output for each input is that we can write the function in a special way. For example, you know about the xy or Cartesian coordinate system, so you already know that you write an equation for a parabola as \(y=x^2\). Notice that the input value, also called the independent variable x and output value y, called the dependent variable since it's value depends on x are written on separate sides of the equal sign. Notice also that there is one and only one y by itself. This is significant in that only functions can be written like this, i.e. with the equation solved for y.

Since we write the y by itself on one side of the equal sign, we can use a special type of notation, which we call function notation. This allows us to drop the y and replace it with a 'name'. This name is most often just a single letter but it can also be longer.

For example, we can rewrite the parabola equation \(y=x^2\) as the function \(f(x)=x^2\). Then we can refer to this parabola equation as the function f(x) or just f. The special notation on the left side of this equation uses parentheses but in this case we are not multiplying f by x. We are saying that the function f has independent variable x. We usually say 'f of x' when referring to this function.

Another nice thing is that we are not tied in to always using f and x when describing this function. The following functions are all equivalent since we always get the same output value associated with each input value.

\(f(x)=x^2\)

\(g(x)=x^2\)

\(h(t)=t^2\)

Evaluating A Function

In order to find out what is going on with a function, we will often need to put in values and see what we get out. The notation \(g(x)=x^2\) makes it really easy to see what we need to do. For example, if we need to know that output value for an input value of 3, we would write \(g(3)\). Comparing this to the function \(g(x)=x^2\), it should give you a clue that we take the input value of 3 and put it everywhere there is an x in the function. So, \(g(3)=3^2=9\), i.e. \(g(3)=9\). It's as simple as that.

If there are more x's in the function, you just replace ALL of the x's with the number. For example, if we have a function \(h(t)=t^2+t-6\), then \(h(3)=3^2+3-6=6 ~~ \to h(3)=6\). That's it.

Let's take a minute to watch a quick video that explains this notation and shows more examples. Toward the end of the video he talks about composition of functions, which we discuss next.

PatrickJMT - Function Notation [5min-29secs]

video by PatrickJMT

Okay, let's work some practice exercises.

Practice

For \(f(x)=3x^2-8x+7\), evaluate \(f(-2)\).

Problem Statement

For \(f(x)=3x^2-8x+7\), evaluate \(f(-2)\).

Solution

1623 video

video by Khan Academy

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For \(g(x)=8x+2\), find \(g(0)\), \(g(-5)\) and \(g(2x+3)\).

Problem Statement

For \(g(x)=8x+2\), find \(g(0)\), \(g(-5)\) and \(g(2x+3)\).

Solution

1622 video

video by Khan Academy

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Next

Next up are four videos that go into a lot of detail about functions. Once you are comfortable with the notation and basics above, these videos will give you a good boost with functions. They contain a lot of detail and involve several topics that we cover on other pages on this site, including composite functions and piecewise functions.

Khan Academy - Introduction to functions [9min-33secs]

video by Khan Academy

Khan Academy - Functions (Part 2) [9min-53secs]

video by Khan Academy

Khan Academy - Functions (Part 3) [9min-14secs]

video by Khan Academy

Khan Academy - Functions (Part 4) [6min-7secs]

video by Khan Academy

Really UNDERSTAND Precalculus

Related Topics and Links

external links that might be helpful

SparkNotes: Functions

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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