Polynomials Day 2
Inverse/compositions
Even and odd functions
Synthetic Division
Characteristics
Inverses
MA2A2. Students will explore inverses of functions.
Discuss the characteristics of functions and their inverses, including one-to-oneness, domain, and range.
Determine inverses of linear, quadratic, and power functions and functions of the form , including the use
of restricted domains.
Explore the graphs of functions and their inverses.
Use composition to verify that functions are inverses of each other.
A function f is one-to-one if for each x
in the domain of f there is exactly one
y in the range and no y in the range is
the image of more than one x in the
domain.
A function is not one-to-one if two
different elements in the domain
correspond to the same element in the
range.
Theorem Horizontal Line Test
If horizontal lines intersect the
graph of a function f in at most
one point, then f is one-to-one.
Use the graph to determine whether
2
the function f ( x )  2 x  5 x  1
is one-to-one.
Not one-to-one.
Use the graph to determine whether the
is one-to-one.
function
One-to-one.
The inverse of a one-one
function is obtained by
switching the role of x and y
Letf ( x)  x
3
and g ( x )  x
Find ( f  g )( x)
and
1
3
( g  f )( x)
g is the inverse of
f.
1
f ( x)  x
1
3
Domain of f
Range of f
f
f
Range of f
1
1
Domain of f
Domain of f  Range of f
1
Range of f  Domain of f
1
1
Theorem
The graph of a function f and the
1
graph of its inverse f are symmetric
with respect to the line y = x.
y=x
f
6
4
f
(0, 2)
1
2
(2, 0)
2
0
2
2
4
6
Finding the inverse of a 1-1
function
Step1: Write the equation in the form
y  f (x)
Step2: Interchange x and y.
Step 3: Solve for y.
Step 4: Write f 1 ( x) for y.
5
Find the inverse of f ( x) 
x 3
5
Step1: y 
x3
Step2: Interchange x and y
Step 3: Solve for y
5
x
y 3
5  3x
f ( x) 
x
1
Even and Odd
Functions
MA2A3. Students will analyze graphs of polynomial functions of higher degree.
b. Understand the effects of the following on the graph of a polynomial function: degree, lead
coefficient, and multiplicity of real zeros.
c. Determine whether a polynomial function has symmetry and whether it is even, odd, or neither.
d. Investigate and explain characteristics of polynomial functions, including domain and range,
intercepts, zeros, relative and absolute extrema, intervals of increase and decrease, and end behavior.
Even functions
A function f is an even function if
f ( x)  f ( x)
for all values of x in the domain of f.
Example: f ( x)  3x 2  1 is even because
f ( x)  3( x) 2  1  3x 2  1  f ( x)
Odd functions
A function f is an odd function if
f ( x)   f ( x)
for all values of x in the domain of f.
Example:f ( x)  5 x 3  x is odd because
f ( x)  5( x)3  x  5x3  x  (5x3  x)   f ( x)
1) f ( x )  x  x  1
4
2) g ( x )  x
2
3
3)h( x) | x | 1
4) k ( x )  x 3  x 2
Determine if the given functions are
even or odd
The graph of an even function is symmetric
with respect to the y-axis.
The graph of an odd function is symmetric
with respect to the origin.
Graphs of Even and Odd
functions
x
f ( x )  4  x 2 g ( x)  x 3  2 x
0
4
0
1
3
-1
-1
3
1
2
0
4
-2
0
-4
y
6
5
4
3
2
1
0
-3
-2
-1
0
-1
-2
-3
-4
-5
-6
1
2
3
x
Determine if the function is even or
odd?
y
5
2.5
0
-5
-2.5
0
2.5
5
x
-2.5
-5
Determine if the function is even or
odd?
y
5
3.75
2.5
1.25
0
-5
-2.5
0
2.5
5
x
Determine if the function is even or
odd?
y
100
50
0
-5
-2.5
0
2.5
5
x
-50
-100
Synthetic Division
MA2A3. Students will analyze graphs of polynomial functions of higher degree.
b. Understand the effects of the following on the graph of a polynomial function: degree, lead
coefficient, and multiplicity of real zeros.
c. Determine whether a polynomial function has symmetry and whether it is even, odd, or neither.
d. Investigate and explain characteristics of polynomial functions, including domain and range,
intercepts, zeros, relative and absolute extrema, intervals of increase and decrease, and end behavior.
3x  x  4x  1
x2  1
3
3x
2
3x 3
 3x
2
x
x 2  1 3x 3  x 2  4x  1
3x
3x 3
x2  x  1
1
x 2  1 3x 3  x 2  4x  1
3x
3x 3
x2  x  1
3x
x2
 1
2
x
x2
1
x
x
3x  1  2
x 1
x 4  2x 2  x  3
x2  x  1
x4
2

x
x2
x3
 x
2
x
x2
x 2  x  1 x 4  0x 3  2x 2  x  3
x 4  x3  x 2
x3  x 2  x
x 2 x
x 2  x  1 x 4  0x 3  2x 2  x  3
x 4  x3  x 2
x3  x 2  x
x3 x2 x
x2  x
x2
x  2 x3  0x2  x  2
x3 2x 2
2x 2  x  2
x3  x  2
x2
x3
 x2
x
x2 2x 5
x  2 x 3  0x 2  x  2
x3 2x 2
2x 2  x  2
2x 2 4x
5x  2
5x  10
2x 2
 2x
x
5x
5
x
12
12
x  2x  5 
x2
2
x 2 x 1
x  3 x 3  4x 2  2x  5
x3  3x 2
x2  2x  5
x 3  4x 2  2x  5
x3
x3
 x2
x
 x 2  3x
x2
 x
x
x
 1
x
x  5
x  3
8
8
x  x  1
x3
2
2x 2 3x 1
x  1 2x 3  x 2  2x  3
2x3  2x 2
3x 2  2x  3
2x 3  x 2  2x  3
x 1
2x 3
 2x 2
x
3x 2  3x
3x 2
 3x
x
x
 1
x
x  3
x  1
4
4
2x  3x  1 
x 1
2
Synthetic Division Summary
1.
2.
3.
4.
Set denominator = 0 and solve (box number)
Bring down first number
Multiply by box number and add until finished
Remainder goes over divisor
Notes of Caution
1.
2.
ALL terms must be represented (even if coefficient is 0)
If box number is a fraction, must divide final answer by
the denominator
To evaluate a function at a particular value, you may EITHER:
A) Substitute the value and simplify OR
B) Complete synthetic division…the remainder is your answer
x  4x  2x  5
x3
3
2
3
x 1 0
x 1
-4
1
x3  0
x3
2x 3  x 2  2x  3
x 1
1
2
-5
3
-3
-3
-1
-1
-8
8
x  x  1
x3
2
1
2
2
-1
2
-3
2
1
3
1
3
0
2x 2  x  3
4x  3x  8x  4
x3
3
2
3
4
x3  0
x3
2x 3  5x 2  28x  14
x5
x5  0
x5
4
-3
-8
4
12
27
57
9
19
61
61
4x  9x  19 
x3
2
5
2
-5
2
-28
14
10
25
-15
5
-3
-1
1
2x  5x  3 
x5
2
2
16x 3  32x 2  81x  162
x2
16 -32
32
0
16
x20
x2
x 3  2x 2  x  1
x3
x3  0
x3
-81
162
0 -162
-81
0
16x 2  81
3
1
1
-2
-1
1
3
3
6
1
2
7
7
x x2
x5
2
x 3  5x  2
x3
3
x3  0x2  5x  2
x3
x 4  0 x 3  17x 2  0x  16
x4
x40
x4
0
1
x3  0
x3
x 4  17x 2  16
x4
1
-5
2
3
9
12
3
4
14
14
x  3x  4 
x3
2
4
1
1
0
-17
0
4
16
-4 -16
4
-1
-4
x3  4x2  x  4
16
0
6x 3  4x 2  3x  2
3x  2
3x  2  0
2
x
3
2x 4  5x 3  4x 2  5x  2
2x  1
2x  1  0
1
x
2
2/3
6
6
-4
3
-2
4
0
2
0
3
0
3
2x 2  1
-1/2 2
2
5
4
5
2
-1
-2
-1
-2
4
2
4
0
2
x3  2x 2  x  2
4x 3  x 2  4x  1
4x  1
4x  1  0
1
x
4
1/4
4
4
-1
-4
1
1
0
-1
0
-4
0
4
x2  1
 1
Find f   if f  x   4x 3  x 2  4x  1
4
 1
f  0
4
4x 3  13x  6
2x  1
2x  1  0
1
x
2
-1/2 4
4
0
-13
-6
-2
1
6
-2
-12
0
2
2x 2  x  6
 1 
Find f   if f  x   4x 3  13x  6
 2
 1 
f  0
 2
x 3  5x  2
x3
x 3  0x 2  5x  2
x3
x3  0
x3
3
1
0
1
-5
2
3
9
12
3
4
14
14
x  3x  4 
x3
2
Find f  3  if f  x   x3  5x  2
f  3   14
x  17x  16
x4
4
2
x 4  0x 3  17x 2  0x  16
x4
x40
x4
4
1
1
0
-17
0
4
16
-4 -16
4
-1
-4
x3  4x2  x  4
Find f  4  if f  x   x 4  17x 2  16
f  4  0
16
0
Direct Substitution

A fancy term for plug in and find
f(x).
Homework- Math 3 Book

Page 69
◦ 1-19 all, skip 8 – 10 and don’t do
part e on 17-19.