Download Properties of Functions New Functions from Old

Properties of Functions
New Functions from Old
Problem. Use the graph to answer these questions about the
function y = f(x).
1. For what values of x is y = 0?
2. For what values of x is y≤ 0 ?
3. What are the maximum and minimum values of y and for
what x do they occur?
4. For what values of x is y = 20?
Problem. Use the equation y = x 2 −3x + 2 to answer these
questions about the function y = f(x).
1. For what values of x is y = 0?
2. For what values of x is y≥ 0 ?
3. Does y have a maximum and minimum value? If so find
them and the x for which they occur.
4. For what values of x is y = −5?
Solution: x2 −3x + 2 = (x −1)(x − 2) so we see first that y = 0
when x = 1 or x = 2. Also, the product of two numbers is >0
when the numbers are both positive or both negative. In this
case both factors are positive if x >2, and both are negative
when x < 1. Thus the answer to 2 is x ≤ 1 or x ≥ 2.
To answer 3, we perform a computation called completing the
square. We know that (x − a)2 = x2 − 2ax + a2 . Thus we can write
x2 −3x + 2 = (x2 −3x + 9) + (2− 9) = (x − 3)2 − 1
4
4
2
4
Since the first factor has a minimum value of 0, we see that the
function is a minimum when x = 3/2, and this minimum value is
–1/4. There is no maximum value, since y gets arbitrarily large
when x does. Finally if we try to solve y = −5, or x2 −3x + 7 = 0
we get no real solutions. Thus this never happens.
Functions from Applied Problem
Often the form of a function must be found from information
describing a practical problem. This problem may also effect
the domain of the resulting function.
Example 8. A box is made from a 6 inch square of cardboard
by cutting out a square of side x from each corner, and folding
up the sides.
x
x
x
x
6
x
x
x
x
V = x(6− 2x)2
x
0≤x≤3
6 −2x
Here the natural domain is not appropriate. We use the
domain that corresponds to the practically realizable values
of the independent variable.
One practical problem is to find the value of x so that the
resulting box has the largest possible volume. We begin by
graphing the function.
It appears from this graph that the maximum volume
occurs when x is about 1. Later in the course we will
be able to prove that it occurs exactly at x = 1.
Piecewise Defined Functions
Sometimes we can combine the methods of specifying a function.
One example that occurs frequently is the piecewise defined
function. Here we describe the function by different formulas that
are valid for different sets of input values. This is, in a way, a
combination of the verbal description and the formula
description.
Example 7. Define a function with real inputs and outputs as
follows:
x2
f(x) =
| x<–1
x+2 | – 1 ≤ x < 3
14- x 2 | x ≥ 3
Then we see that
f (−2) = (−2)2 = 4
f (0) = 0 + 2 = 2 and
f (4) =14 −(4)2 = −2
The graph of this function is:
One of the most useful functions defined in this way is the
absolute value function. We let
|x| =
x| x≥0
–x| x<0
The graph of the function whose formula is f(x) = |x| is :
We see that the expression |x – a| will equal x – a when x − a ≥0
and will equal –(x – a) when (x – a)<0. In other words
x–a| x≥a
|x – a| =
a–x| x<a
Thus we can simplify the following functions:
5x – 2 | x ≥ 0
1. |x| + 4x – 2 =
3x – 2 | x < 0
3x + 1 | x ≥ 1/3
2. 2 + |3x – 1| =
– 3x + 3 | x < 1/3
Since 3x – 1 ≥ 0 means that 3x ≥ 1, or x ≥ 1/3.
2x – 4| x ≥ 3
3. |x – 1| + |x – 3| =
2 | 1 ≤ x<3
4 – 2x | x< 1
Since the expression becomes respectively (x – 1) + (x – 3),
(x – 1) + (3 – x), and (1 – x) + (3 – x)
New Functions from Old
Given any two numerical functions, we can combine these
functions by addition, subtraction, multiplication and division
to get new functions.
Let f and g be two numerical functions. Then we define f + g,
f − g, fg, and f/g to be the functions whose values are produced
by the following formulas:
1. (f + g) (x) = f(x) + g(x)
2. (f − g) (x) = f(x) − g(x)
3. (fg)(x) = f(x)g(x)
4. (f/g)(x) = f(x)/g(x)
The domains of the first three are taken to be the intersection
of the domains of f and g (all numbers common to both), and
for 4. We also exclude any place where g(x) = 0.
x
f
g
f+g
f(x)
g(x)
y = (f + g)(x)
= f(x) + g(x)
Example 1. Let f and g be defined by the formulas
f(x) = 2 + x −1
g(x) = x− 4
and let each have their natural domain. Then
1. (f + g) (x) = f(x) + g(x) = 2 +
x−1 + (x − 4) = (x − 2) +
x−1
2. (f − g) (x) = f(x) − g(x) = 2 + x−1 − (x − 4) = (−x + 6) + x−1
3. (fg)(x) = f(x)g(x) = (2 + x −1) (x − 4)
4. (f/g)(x) = f(x)/g(x) = (2 + x −1)
x−4
The domain natural of the first three is all x≥1
In the case of number 4, the natural domain consists of all
x≥1 except x = 4.
1
Example 2. Let f(x) = x and g(x) = 2 .
x
The natural domain of f is all non negative numbers, and that of g
is all real numbers other than 0. The common domain is therefore
all positive real numbers.
For x positive, (f + g) (x) = f(x) + g(x) =
shown below.
x +
1
2 . The graphs are
x
x + 12
x
x
1
x2
Stretches and Compressions
If we multiply a function by a positive constant, we get a new
function whose graph is the graph of the original function
stretched or compressed vertically, depending on whether the
constant is greater than or less than 1.
If we multiply the independent variable by a positive constant,
the new function has a graph that is the graph of the original
function stretched or compressed horizontally, again depending
on whether the constant is less than or greater than 1.
Let f be the function whose graph is shown below:
We combine this function with 2f and .5f on the same graph.
f
2f
.5f
Now we combine the functions with formulas f (x), f(2x),
and f (.5x) on the same graph.
.5x
f(2x)
f (x)
Warning. Multiplying the independent variable by numbers
> 1 compresses the graph. Multiplying by numbers < 1
stretches it.
Translations
If we add a nonzero constant to a function the graph is
translated up or down by that amount, depending on whether
the constant is positive or negative.
If we add a nonzero constant to the independent variable, the
graph is translated right or left, depending on whether the
constant is negative or positive.
Warning. Note that in the last statement, the sign of the
constant and the direction of the horizontal shift are
opposite to one another.
Let f(x) = |x|. We have the following graphs:
|x − 2|
|x| + 2|
|x + 2|
|x| −2
Symmetry and reflections
If we multiply a function by –1, we reflect the graph about the x
axis. If we multiply the independent variable by –1, we reflect
the graph about the y axis.
f(x)
f(x)
f(– x)
– f(x)