FUNCTION
A function is a set of ordered pairs which relates two sets such that to each element of one set there is only one element of the second set
Example
Represent the following set of ordered pairs in a pictorial diagram.
(1,4), (2,3), (-1,2), (-2,-3).
Solution
(1,4), (2,3), (-1,2), (-2,-3).
Solution
A function whose graph is such that any line drawn parallel to the x – axis at any point cuts it at only one point is one-to-one function.
Examples.
1. Write the expression of a function ‘ double plus one’
Solution
f:x 
2x +1
2x +1b)
2.Given the function G(x) = 4x-1. Find the value of G(-2).
Solution
2.Given the function G(x) = 4x-1. Find the value of G(-2).
Solution
G(x) = 4x -1
G(-2) = 4(-2) -1
G(-2) = -8 -1
G (-2) = -9
Exercise 2.1
1. Write each of the following function in the form f: x f(x) use any functions symbol to represent the functions
f(x) use any functions symbol to represent the functions(a)Divide by 5 and add 2.
(b)Subtract 7 and square
(b)Cube and then double
(b)Subtract 7 and square
(b)Cube and then double
Solution
(a) F:x f(x)
f(x) F:x + 2
+ 2 (b) F: x (x-7)2
(x-7)2 (c) F:x x3+2x
x3+2x2. Find the value of the function for each given value of x.
(a) f(x)= 2x+3; when
(i) x=1
(ii) x= -2
(iii) x =a
(i) x=1
(ii) x= -2
(iii) x =a
Solution
(a) (i) when x=1
f(x)= 2x+3
f(1)= 2(1)+3
f(1)= 2+3
∴ f(1)= 5.
(ii)when x= -2.
f(x) = 2(-2) +3
f(-2) = -4+3
∴f(-2) = -1
(iii) when x=a
f(x) = 2(a) +3
f(a) =2a+3
∴f(a) = 2a+3
(b) C(x) = x3 when
(i) x=1
(ii) x = -1
(iii)x= 0
(iv)x=b
Solution
(i) C(x) =x3
C (1) =13
C (1) =1
C (1) =1
(ii)C(x) =x3
C (-1) = (-1)3
C (-1) = -1
(iii)C(x) =x3
(iii)C(x) =x3
C (0) =03
C (0) =0
(iv)C(x) =x3
C(b) = b3
(c) K(x) = 3-x; when
(i) x= -1
(ii)x=7
Solution
(i) K(x) = 3-x
K (-1) = 3- (-1)
K (-1) = 4
(ii) k(x)=3 – x
k(7) = 3-7
k(7) = -4
(ii) k(x)=3 – x
k(7) = 3-7
k(7) = -4
DOMAIN AND RANGE OF FUNCTIONS
Example 1.
1. Find the domain and range of f(x) =2x+1
let f(x) = y
y=2x+1
y=2x+1
Solution
Domain = {x: xR}
R} Range – make x the subject
y=2x+1
y-1=2x
(y-1)/2 =x
∴x=(y-1)/2
Range={y: yR}
y=2x+1
y-1=2x
(y-1)/2 =x
∴x=(y-1)/2
Range={y: y
R}b Example 2
If y =4x+7 and its domain ={x : -10 < x< 10} find the range.
If y =4x+7 and its domain ={x : -10 < x< 10} find the range.
Solution
Domain = {x: -10 < x< 10}
Range ={y : -33 < y < 47}
y = 4x + 7
Table of values
Table of values
c Example 3.
Y= √x and domain is -5 < x < 5, Find its range.
Y= √x and domain is -5 < x < 5, Find its range.
Solution
Domain = {x: -5 < x < 5}
Range =y
Table of value of x = √(y)
(y)2= (√x)2
Table of value of x = √(y)
(y)2= (√x)2
y2= x
x= y2
√5= √y2
y = √5
Range = {y: o < y < √5}
d Example 4.
Given F(x) = find the domain and range.
Given F(x) =
find the domain and range. Solution
Let f(x)=y
Domain of y=
For real values of y: 1-x2 > 0
Domain of y=
For real values of y: 1-x2 > 0
-x2 > 0-1
-x2 > -1
x2 < 1
√x2< √1
X <√1
X <√1
X <1
∴Domain {x : X <1 }
let F(x) = y
y=
To get the range, make x the subject
y2 = ()2
)2 y2= 1-x2
X2=1-y2
Therefore =
= 
X=
For real value of x:
1- y2 > 0
y2 < 1
y < √1
1- y2 > 0
y2 < 1
y < √1
y < 1
...Range ={y : y <1}
LINEAR FUNCTIONS
Is the function with form f(x) = mx + c.
Where:
Where:
f(x)= y
m and c are real numbers
m is called gradient[ slope].
c is called y – intercept.
Example
1. Find the linear function f(x) given the slope of -2 and f( -1)=3
Solution
Given:
m(slope)=-2
x=-1
f(-1)=3
Given:
m(slope)=-2
x=-1
f(-1)=3
from;
f(x) =mx +c
f(x) =mx +c
f(-1) =( -2x-1)+c
3 =2+c
3-2 =c
c=1
∴f(x) = -2x+1
Example 2.
Find the linear function f(x) when m=3 and it passes through the points (2, 1)
Solution
f(x) =mx +c
f(2) = 3(2)+ c
1= 6+c
1-6 =c
-5 = c
c= -5
f(x) = 3x + -5
∴ f(x) =3x-5
Example 3
Find the linear function f (x) which passes through the points (-1, 1) and (0, 2 )
Solution
Slope =
=1
m = 1
f(x) = mx +c
f(-1) = 1x(-1) +c
1 =-1+c
c=2
c=2
f(x) = 1x +2
∴f(x) = x +2.
4.Draw the graph of h(x) = 3x-4
4.Draw the graph of h(x) = 3x-4
Table of values of function
X | -1 | 0 | 1 |
h(x) | -7 | -4 | -1 |
Exercise
In problems 1 to 3 find the equation of a linear function f(x) which satisfies the given properties. In each case, m dissolves the gradient.
1). m =-3, f(1) = 3
2). m=2, f(0) =5
3). f(1) =2, f(-1) =3
4. Givenm= -4 , f(3) = -4 Find f(x)
In the problem 5 to 9 draw the graphs of each of the given functions without using the table of values
5) f(x)=+
6) f(x) =4
Solution
2). m=2, f(0) =5
3). f(1) =2, f(-1) =3
4. Givenm= -4 , f(3) = -4 Find f(x)
In the problem 5 to 9 draw the graphs of each of the given functions without using the table of values
5) f(x)=
+
6) f(x) =4
Solution
1. f(x) = mx + c
f(x) = -3(1)+c
f(x) =-3 +c
3= -3 +c
3+3 =c
6=c
C=6
f(x) = -3x +6
2. f(x) =mx +c
f(0) = (2 x 0)+c
5= 0+c
c=5
f(x) =2x+5
3)
3.f(1) =2, f(-1) =3 Alternatively
3.f(1) =2, f(-1) =3 Alternatively
m = f(-1) = m(-1) + c
f(-1) = m(-1) + c M= 3= -m + c…………………..(i)
3= -m + c…………………..(i) f(1)=2 f(1) = m(1) + c
f(x)= mx +c 2= m + c(ii)
f(x)= mx +c 2= m + c(ii)
f(1)= –x 1+c Solve (i) and (ii) Simultaneously
x 1+c Solve (i) and (ii) Simultaneously f(1) = –+c -m + c = 3
+c -m + c = 3 2= +c + m + c = 2
+c + m + c = 2 2+= c C=5/2
= c C=5/2 C=2 put c in (i)
put c in (i) f(x) =+ 2 3 = -m + 5/2
+ 2 3 = -m + 5/2 m = 5/2 – 3
m= -1/2
...f(x) =+ 2
4.
m= -1/2
...f(x) =
+ 2 4.
f(x)= mx + c
f(x) = -4(3)+c
-4 =-12+c
-4+12= c
8=c
C=8
f(x) = -4x+8
5. f(x)=
+ y- intercept, x=0
f(0)= 2/5[0] +1/5
f(0)= 0+1/5
y=1/5
{0,0.2}
x- intercept, y=0
0=2/5[x] +1/5
x intercept = – ½ ( -1/2, 0)
y intercept = 1/5 ( 0,0.2)
6. f(x) =4
From f(x) = y
From f(x) = y
y = 4
QUADRATIC FUNCTIONS
A quadratic function is any function of the form;
f(x)=ax2+bx+c
Where a ≠ 0
a ,b and c are real numbers.
When a = 1, b=0, and c= 0
X | -3 | -2 | -1 | 1 | 2 | 3 | |
f(x) | 9 | 4 | 1 | 0 | 1 | 4 | 9 |
The shape of the graph of f(x)=ax2+bx+c is a parabola
•The line that divides the curve into two equal parts is called a line of symmetry [axis of symmetry]
•Point (0,0) in f(x)=x2 called the turning point (vertex).
If “a” is positive the turning point is called minimum point (least value).
If “a” is negative the turning point is called maximum point.
PROPERTIES OF QUADRATIC FUNCTIONS
Think of f(x)= ax2+bx+c
y=a(x2+bx/a)+c
y= a( x2+bx/a+b2/4a2)+ c-b2/4a2
=a(x + ) 2 +
) 2 +
x> 0 then
a(x+)2 > 0
)2 > 0 y =
This is when x = –
The turning point of the quadratic function is )
)Example
Find the minimum or maximum point and line of symmetry f(x) = x2 – 2x-3. Draw the graph of f(x)
Solution:
Turning point = )
) = ( – )
)Maximum point = (1, -4)
Line of symmetry is x=1
x | -5 | -2 | -1 | 0 | 1 | 2 | 4 | |
f(x) | 15 | 5 | 0 | -3 | -4 | -3 | 0 | 5 |
Exercise
1. Draw the graph of the function y=x2-6x+5 find the least value of this function and the corresponding value of x
Solution
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
y | 32 | 21 | 12 | 5 | -3 | -4 | -3 | 0 |
Least value.
y= – 4 where x=3
2. Draw the graph of the function y =x2-4x+2 find the maximum function and the corresponding value of x use the curve to solve the following equations
a) X2-4x-2 = 0
b) X2-4x-2 = 3
a) X2-4x-2 = 0
b) X2-4x-2 = 3
b)
Solution
Table of values of y= x2-4x+2
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
y | 23 | 7 | 2 | -1 | 2 | -1 | 2 | 7 | 14 |
Find the maximum value of the function.;
Maximum value
=
Y= ( ) = – 2
) = – 2Find the maximum value of the function;
Maximum value= )
) Maximum value= )
) Maximum value
[2, -2]
The maximum value is = (-2,-2)
a) x2-4x-2 = 0
add 4 both sides
x2-4x + 2 = 4
but x2-4x+2 = y
... y = 4
Draw a line y = 4 to the graph above. The solution from the graph is
x1 = -1/2 , x2 = 9/2
(x1,x2) = (-1/2,9/2 )
(b) x2-4x – 2 = 3
a) x2-4x-2 = 0
add 4 both sides
x2-4x + 2 = 4
but x2-4x+2 = y
... y = 4
Draw a line y = 4 to the graph above. The solution from the graph is
x1 = -1/2 , x2 = 9/2
(x1,x2) = (-1/2,9/2 )
(b) x2-4x – 2 = 3
add 4 both sides
x2-4x – 2 +4 = 3+ 4
x2-4x + 2 = 7
but x2-4x+2 = y
... y = 7
Draw a line y = 7 to the graph above. The solution from the graph is
x1 = -1 , x2 = 5
(x1,x2) = (-1,5 )
3. In the problem 3 to 5 write the function in the form f(x)= a( x + b)2 +c where a, b, c are constants
x2-4x – 2 +4 = 3+ 4
x2-4x + 2 = 7
but x2-4x+2 = y
... y = 7
Draw a line y = 7 to the graph above. The solution from the graph is
x1 = -1 , x2 = 5
(x1,x2) = (-1,5 )
3. In the problem 3 to 5 write the function in the form f(x)= a( x + b)2 +c where a, b, c are constants
f(x) =5-x-9x2
Solution
f(x) = -9x2 – x + 5
= -9( x2 – ) + 5
) + 5 = -9( x2 – + ) + 5 +
+ 
) + 5 + 
= -9(x – )2 +
)2 + 
4. In the following functions find:
a) The maximum value
b) The axis of the symmetry
f(x)= x2– 8x+18
Solution
) = 
) ( 4, 2)
Maximum value =2 where the axis of symmetry x=4.
5. f(x) = 2x2+3x+1
5. f(x) = 2x2+3x+1
Solution
Maximum value )
) 
) The turning point of the graph is
The minimum value of the graph is y = –
axis of symmetry =
POLYNOMIAL FUNCTIONS
axis of symmetry =
POLYNOMIAL FUNCTIONS
The polynomial functions are the functions of the form, ” P(x) =an xn + a n-1 xn-1 + an-2 xn-2 + . . . + a1 x1+a0 x0 “. Where n is non negative integer and an, an-1, an-2 . . . a0 are real numbers. The degree of a polynomial function is the highest power of that polynomial function.
Example
a) f(x) =5x4 -7x3 +8x2– 2x+3 is a degree of 4
b) H(x)=6x-8x2+9x9-6 is a degree of 9
c) G(x) =16x-7 is a degree of 1
d) M(x) =6 degree is 0 =6x0
GRAPHS OF POLYNOMIAL FUNCTIONS
EXAMPLE
Draw the graph of f(x) = x3-2x2-5x+6
Solution
Table of values
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 | |
F[x] | -24 | 0 | 8 | 6 | 0 | -4 | 0 | 18 |
STEP FUNCTIONS
EXAMPLE
1. If f is a function such that;
Draw the graph and find its domain and range.
Domain = {x: xR, except -3< x < -2}
R, except -3< x < -2}Range = {1,4, -2}
2. The function is defined by
a)
Sketch the graph of f(x) use the graph to determine the range and the domain. Find the value of f(-6) , f(0). State if it is a one to one function
Sketch the graph of f(x) use the graph to determine the range and the domain. Find the value of f(-6) , f(0). State if it is a one to one function
Domain
= {x : XR}
R} Range
{ y:y >1}
f(-6) = -2
f(0)= 2
It is not a one to one function.
EXERCISE
1. Draw the graph of the following defined as indicated
Solution:
Solution
2. Given that f(x) =
a) On the same set of axes sketch the graphs of f(x )and the inverse of f(x), From your graphs in [a] above determine;
∙ (a)The domain and range of f(x)
(b)The domain and range if the inverse of f(x)
(c)Find f(- 5)and f(5)
(d)Is f (x)a one to one?
(e)Is the inverse of f(x) a function?
(b)The domain and range if the inverse of f(x)
(c)Find f(- 5)and f(5)
(d)Is f (x)a one to one?
(e)Is the inverse of f(x) a function?
(a) Domain of f(x ) = { x: x }
}(b) •Range of f(x) = { y : y }
•Domain of f-1(x ) = { x : x}
}•Domain of f-1(x ) = { x : x
}(c) Range of f-1(x ) = { y: x }
}(d) f(-5) = 1 and f(5) = 6
(e)Yes the inverse of f(x) is a function and f(x) is one-to-one function
ABSOLUTE VALUE FUNCTIONS
The absolute value function is defined by f(x)
Example
1. Draw the graph of f(x) = | x |
Solution
Table of values f(x) = | x |
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
f(x) | 3 | 2 | 1 | 0 | 1 | 2 | 3 |
THE INVERSE OF A FUNCTION
Given a functon y= f(x), the inverse of f(x) is denoted as f-1(x). The inverse of a function can be obtained by interchanging y with x (interchanging variables) and then make y the subject of the formula.
Example
1. Find the inverse of f(x) = 2x+3
Solution
Y= 2x+3
X=2y+3
2y=x-3
Y =
∴ f-1(x
EXPONENTIAL FUNCTIONS
An exponential function is the function of the form f(x) = nx where n is called base and x is called exponent.
Example
1. Draw the graph of f(x)=2x
1. Draw the graph of f(x)=2x
Solution :
Table of values
X | -3 | -2 | -1 | 0 | 1 | 3 | |
f(x) | 1/8 | 1/4 | ½ | 1 | 2 | 4 | 8 |
2. Draw the graph of f (x)=2-x
Solution:
Table of values
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
f(X) | 8 | 4 | 2 | 1 | 1/2 | ¼ |
Properties of exponential function
- When x increases without bound, the function values increase without bound
- When x decreases, the function values decreases toward zero
- The graph of any exponential function passes through the point (0,1).
- The domain of the exponential function consists of all real numbers whereas the range consist of all positive values.
LOGARITHMIC FUNCTION
Logarithmic function is any function of the form f(x)= read as function of logarithm x under base a or f(x) is the logarithm x base a
read as function of logarithm x under base a or f(x) is the logarithm x base aExample
Draw the graph of f(x)=
Solution
Table of values
x | 1/8 | ¼ | 1/2 | 1 | 2 | 4 | 8 |
f(x) | -3 | -2 | -1 | 0 | 1 | 2 |
THE INVERSE OF EXPONENTIAL AND LOGARITHMIC FUNCTION
The inverse of the exponential function is the relation of logarithmic in the line y=x
EXAMPLE
1. Draw the graph of the inverse of f(x) = 2x and f(x)=under the same graph.
Solution
1. Draw the graph of the inverse of f(x) = 2x and f(x)=
under the same graph.Solution
(i) y= 2x
Apply log on both sides
log x=log 2y
log x =y log 2
y = log (x- 2)
f-1(x) = log (x- 2)
(ii) f(x)=
y=
x=
log x=log 2y
log x =y log 2
y = log (x- 2)
f-1(x) = log (x- 2)
(ii) f(x)=
y=
x=
y= 2x
Table of values
X | 1/8 | 1/4 | 1/2 | 2 | 4 | 8 | |
f(x) | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
2. Draw the graph of the function f(x)=3x
Table of values
x | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
f(X) | 1/27 | 1/9 | 1/3 | 1 | 3 | 9 | 27 |
3. Draw the graph of the function f(X)=8x
Solution
Table of values
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
f(X) | 1/512 | 1/64 | 1 | 8 | 64 | 512 |
Exercise
1. Find the graph of y =2x and given that ¾ =2-0.42 draw the graph of f(x)= (3/4)x
Table of values if f(x) =(3/4)x
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
2x | 1/8 | 1/4 | ½ | 1 | 2 | 4 | 8 |
-0.42x | 1.26 | 0.42 | 0 | -0.42 | -0.84 | 1.26 | |
2-0.42x | 64/27 | 16/9 | 4/3 | 1 | 3/4 | 9/16 | 27/64 |
Copy and complete the following table and hence draw the graph of f(x)=(1/4)x
X | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
2x | -6 | -4 | -2 | 0 | 2 | 4 | |
-2x | 6 | 4 | 2 | 0 | -2 | -4 | -6 |
2-2x | 64 | 16 | 4 | 0 | 1/4 | 1/16 | 1/64 |
[1/4]x | 64 | 16 | 4 | 0 | 1/4 | 2/4 | ¾ |
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