The document provides an overview of polynomial functions, their characteristics including degrees and leading coefficients, and the smooth and continuous nature of their graphs. It also covers absolute value equations and the methods for solving these equations both algebraically and graphically, as well as the definition and significance of one-to-one functions and their inverses. Additionally, it discusses algebraic functions and operations with polynomials, such as function combinations and compositions.

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A polynomial function in one variable is a function of the form
f (x)= anxn
+ an-1xn-1
+….. + a1x + a0
where an , an-1 , a1 , a0 are constants, called the coefficients of the polynomial, n ≥
0 is an integer, and x is the variable. If an ≠ 0, it is called the leading coefficient,
and n is the degree of the polynomial. The domain of a polynomial function is the
set of all real numbers.

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The number an, the coefficient of the highest power is the leading coefficient, and the
term anxn is the leading term.
We often refer to polynomial functions simply as polynomials. The following
polynomial has degree 5, leading coefficient 3, and constant term -6.

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the graph of every polynomial function is both smooth and continuous. By smooth, we
mean that the graph contains no sharp corners or cusps; by continuous, we mean that
the graph has no gaps or holes and can be drawn without lifting your pencil from the
paper. See Figures (a) and (b).

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A power function of degree n is a monomial function of the form f(x) = axn
where a is a real number, a ≠ 0, and n < 0 is an integer
Properties of Power Functions, f(x) = xn
, n Is a Positive Even Integer
1. f is an even function, so its graph is symmetric with respect to the y-axis.
2. The domain is the set of all real numbers. The range is the set of
nonnegative real numbers.
3. The graph always contains the points (-1 , 1), (0,0) and (1,1).
4. As the exponent n increases in magnitude, the graph is steeper when
x > - 1 or x < 1; but for x near the origin, the graph tends to flatten out and lie
closer to the x-axis

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Absolute Value Equations:
Definition of the absolute value
If x is a real variable, a, b, are real numbers,
then
And so
And

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Properties of the absolute value of the real number
The absolute value of the sum of two numbers is smaller than or equal to the sum of
their absolute values and the equality is happened if a , b are negative together ,
positive together or each of them equals zero.
Remarks:
1. for any real number a, then:
2.
3.
4. If a and b are two real numbers, then:
5. For any real number a , then :
6. for any real number a, then:
7. If, then :
8. If, then :

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•Solving the equation in the form :
|aX + b| = c, c [0,∞[
∈
Absolute of first degree expression = non negative real number
•algebraic solution
(1) Using the definition
(2) Using the property value inside the absolute sign = ± the real number
•graphical solution
The X-coordinates of the intersection points of the two curves f(x) = |a x + b│, g (x) = c
•Remark:
If |ax+ b|=c, c ]-∞,0[ ,then the solution set in R = Ø
∈
For example, the solution set of the equation |3 x-4| = - 5 in R is Ø

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•Example (1):
Find graphically, then perform algebraically the solution set in R for the following equation:
|X-2| = 3
•Solution: Graphically:
Putting f (X) = |x-2|, g (X) = 3
• We draw the curve of the function f: f (x) = |x-2| and it is the same curve of y = |X| with a
horizontal translation 2 units in the direction OX
• We draw the curve of the function g: g (X) = 3 and it is a constant function
represented by a straight line parallel to the X-axis and intersects
the y-axis at the point (0,3)
• We find the intersection points of the two curves are (-1, 3) and (5,3)
∴The solution set = {-1,5}

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Algebraically:
First: Using the definition of the absolute value function
At
At
The solution set = {-1,5}
Second : Using the property "what inside the absolute sign = ± the real number"

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•Solving the equation in the form:
|aX + b| = |cX + d|
Absolute of first degree expression in X = absolute of first degree expression in X
algebraic solution
(1) One of the two expressions = ± the other expression.
(2) By squaring the two sides of the equation.
graphical solution
The X-coordinates of the intersection points of the two curves f(x) =│ax+b│, g (x) = │cx + d │
Example (2):
Find graphically, then perform algebraically the solution set of the equation :
|x-4| = |2x-5│in R
Solution:
Put f(x) = │x-4│ ,g(x) = │2x-5│ = 2│x - 2.5│

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Graphically:
The function f is represented graphically by the curve y = | x | with
horizontal translation 4 units in OX directions; the function g is
represented graphically by the graph y = 2 │ x │ with horizontal
translation 212 units in OX direction.
∴the two curves are intersecting at the two points
(1 , 3) , (3 , 1)
∴The solution set = {1 , 3}

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Algebraically:
First: By using the property "one of the two expressions = ± the other
expression"
(From absolute property)
Second: By squaring both sides

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Inverse of a function:
In order to understand the inverse of a function we must first get to know the term one-to-one
function:
A one-to-one function is when no two numbers in a set have the same image, so if two numbers have
same image it is not a one-to-one function
In f function any two numbers in A have different images,
but in g function both 2 and 3 have the same image, 4.
Inverse of a function:
If a one-to-one function has a domain A and a range B, the inverse of this function has a domain B and a
range A, so an inverse function is the inverse to one-to-one function, for example, let f be one-to-one
function & f-1 be the inverse of a function:
f-1y=x fx=y
The arrow diagram indicates that f-1 reverses the
Effect of f, from the definition we have:
Domain of f-1 = range of f
Range of f-1 = domain of f

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Test one-to-one function: (graphically)
Horizontal line test
A function is one-to-one if and only if no horizontal line intersects its graph more than once.
For example:
As shown in the opposite figure there are numbers x1 and x2 such that fx1=f(x2) this means that f is not
one-to-one.
Solution 1: As there is no horizontal line intersects the graph
more than once therefore fx= x3 is one-to-one function.
Solution 2: If x1≠x2 , then x13≠x23 (two different numbers
can’t have the same cube). Therefore fx= x3 is one-to-one.

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applying a function f and then its inverse f-1
gives us the original
value back again:
f-1
( f(x) ) = x
f(x) = 2x+3
Using the formulas from above, we can start with x=4:
f(4) = 2×4+3 = 11
We can then use the inverse on the 11:
f-1
(11) = (11-3)/2 = 4
And we magically get 4 back again!
We can write that in one line:
f-1
( f(4) ) = 4
"f inverse of f of 4 equals 4“
domain of
domain of

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When we square a negative number, and then do the inverse, this happens:
Square:(−2)2
= 4
Inverse (Square Root):√(4) = 2
But we didn't get the original value back! We got 2 instead of −2. Our fault for not
being careful!
So the square function (as it stands) does not have an inverse
But we can fix that!
Restrict the Domain (the values that can go into a function).
Example: (continued)
Just make sure we don't use negative numbers.
In other words, restrict it to x ≥ 0 and then we can have an inverse.
So we have this situation:
•x2
does not have an inverse
•but {x2
| x ≥ 0 } (which says "x squared such that x is greater than or equal to zero"
using set-builder notation) does have an inverse.

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Let us see graphically what is going on here:
To be able to have an inverse we need unique values.
Just think ... if there are two or more x-values for one y-value, how do we know
which one to choose when going back?
Imagine we came from x1
to a particular y value, where do we go back to? x1
or
x2
?
In that case we can't have an inverse.
But if we can have exactly one x for every y we can have an inverse.
It is called a "one-to-one correspondence" or Bijective, like this
The graph of f(x) and f-1
(x) are symmetric across the line y=x

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Algebraic function
An algebraic function is a function that involves only algebraic operations. These
operations include addition, subtraction, multiplication, division, and exponentiation.
Eamples:
•f(x) = x2
- 5x + 7
•g(x) = √x
•h(x) = (3x + 1) / (2x - 1)
•k(x) = x3
Non-algebraic functions include trigonometric functions, logarithmic functions,
absolute value functions, exponential functions, etc. Here are some examples.
•f(x) = sin (3x + 2)
•g(x) = log x
•h(x) = 3x

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Algebraic Functions Graphs
The graphs of all algebraic functions are NOT the same. It depends upon the equation of
the function. The general procedure to graph any y = f(x) is:
•Find the x-intercepts (by setting y = 0)
•Find the y-intercepts (by setting x = 0)
•Find all asymptotes and plot them.
•Find the critical points and inflection points.
•Find some extra points in between every two x-intercepts and in between every two
asymptotes.
•Plot all these points and join them curves by taking care of the asymptotes.

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Operations with polynomials:
Combinations of Functions
Two functions and can be combined to form new functions in a manner similar to the way
we add, subtract, multiply, and divide real numbers.
The sum and difference functions are defined by:
If the domain of is A and the domain of is B, then the domain of is the intersection A B
Ո
because both and have to be defined.
For example:
The domain of = A = [0, ∞)
The domain of = B = (-∞, 2]
So, the domain of is A B = [0, 2].
Ո

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Similarly, the product and quotient functions are defined by:
Fg(x)=f(x)g(x)
The domain of fg is A B.
Ո
Because we can’t divide by 0, the domain of is therefore { x A B│g(x)≠0}
∈ Ո
For example:
If f(x)=x2
and g(x)=x-1
Then the domain of the rational function (x)=x2
(x-1) is
{x│x≠1} or (-∞,1)ꓴ(1,∞)
There is another way of combining two functions to obtain a new function. For example, suppose that y=f(u)= and
the u=g(x)=x2
+1 since y is a function of u and u is in turn a function of x it follows that y is ultimately a function of x.
We compute this by substitution:
y=f(u)=f(g(x))=f(x2
+1)=x2
+1
The procedure is called composition because the new function is composed of the two given functions f and g.

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In general, given any two functions f and g, we start with a number x in the domain of g and
calculate g(x). If this number g(x) is in the domain of f, then we can calculate the value of
f(g(x)). Notice that the output of one function is used as the input to the next function. The
result is a new function h(x) = f(g(x)) obtained by substituting g into f. It is called the
composition (or composite) of f and g and is denoted by fog (“f circle g”).
The domain of fog is the set of all x in the domain of g such that gx is in the domain off. In
other words, fog(x) is defined whenever both g(x) and f(g(x)) are defined.
For example:
If fx=x2 and gx=x-3, find the composite functions fog and gof
(Fog)(x)=fg(x)=f(x-3)=(x-3)2
(Gof)(x)=gf(x)=g(x2
)=(x2
-3)