# 3.1 Functions and function notation  (Page 8/21)

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The most common graphs name the input value $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and the output value $\text{\hspace{0.17em}}y,\text{\hspace{0.17em}}$ and we say $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ is a function of $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}y=f\left(x\right)\text{\hspace{0.17em}}$ when the function is named $\text{\hspace{0.17em}}f.\text{\hspace{0.17em}}$ The graph of the function is the set of all points $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ in the plane that satisfies the equation $y=f\left(x\right).\text{\hspace{0.17em}}$ If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x -coordinate of each point is an input value and the y -coordinate of each point is the corresponding output value. For example, the black dots on the graph in [link] tell us that $\text{\hspace{0.17em}}f\left(0\right)=2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(6\right)=1.\text{\hspace{0.17em}}$ However, the set of all points $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ satisfying $\text{\hspace{0.17em}}y=f\left(x\right)\text{\hspace{0.17em}}$ is a curve. The curve shown includes $\text{\hspace{0.17em}}\left(0,2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(6,1\right)\text{\hspace{0.17em}}$ because the curve passes through those points.

The vertical line test    can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See [link] .

Given a graph, use the vertical line test to determine if the graph represents a function.

1. Inspect the graph to see if any vertical line drawn would intersect the curve more than once.
2. If there is any such line, determine that the graph does not represent a function.

## Applying the vertical line test

Which of the graphs in [link] represent(s) a function $\text{\hspace{0.17em}}y=f\left(x\right)?$

If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of [link] . From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x -values, a vertical line would intersect the graph at more than one point, as shown in [link] .

Does the graph in [link] represent a function?

yes

## Using the horizontal line test

Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test    . Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function.

Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function.

1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once.
2. If there is any such line, determine that the function is not one-to-one.

## Applying the horizontal line test

Consider the functions shown in [link] (a) and [link] (b) . Are either of the functions one-to-one?

The function in [link] (a) is not one-to-one. The horizontal line shown in [link] intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.)

The function in [link] (b) is one-to-one. Any horizontal line will intersect a diagonal line at most once.

Is the graph shown in [link] one-to-one?

No, because it does not pass the horizontal line test.

## Identifying basic toolkit functions

In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ as the input variable and $\text{\hspace{0.17em}}y=f\left(x\right)\text{\hspace{0.17em}}$ as the output variable.

what is math number
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of \$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar