Form 3 Mathematics – 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

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

b)
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

(a)Divide by 5 and add 2.
(b)Subtract 7 and square
(b)Cube and then double

Solution

(a)        F:x            f(x)

            F:x   
  +  2

     (b)       F: x    (x-7)²

     (c)       F:x     x³+2x

2. 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

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

   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

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

  Solution

   Domain = {x: xR}

   Range – make x the subject
y=2x+1
y-1=2x
(y-1)/2 =x
∴x=(y-1)/2
Range={y: yR}

b  Example 2
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

c   Example 3. 

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

  y²= x

  x= y2

  √5= √y²

  y = √5

    Range = {y: o < y < √5}

d  Example 4.
Given F(x) = find the domain and range.

 Solution

Let f(x)=y
Domain of y=  
For real values of y: 1-x²
> 0

   -x²
> 0-1

   -x²
> -1

    x²
< 1

   √x²< √1

X <√1

   X <1

   

  ∴Domain {x : X <1 }

  let F(x) = y

   y=

To get the range, make x the subject

   y² = ()²

   y²= 1-x²

   X²=1-y²

  Therefore = 

   X=

 ^{For real value of x:
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:

   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

  from;
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

     f(x) = 1x +2

    ∴f(x)  = x +2.

4.Draw the graph of h(x) = 3x-4

     Table of values of function 

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

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

       m  =                                      f(-1) = m(-1) + c

        M=                                                                  3= -m + c…………………..(i)

    f(1)=2                                             f(1) = m(1) + c

f(x)= mx +c                                     2= m + c(ii)

    f(1)= ^–x 1+c                                Solve (i) and (ii) Simultaneously

    f(1) = ^–+c                                   -m + c = 3

         2= +c                               + m + c = 2

         2+= c                                              C=5/2

         C=2                                                                        put c in (i)

    f(x) =+ 2                                             3 = -m + 5/2

                                                         m = 5/2 – 3
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

   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

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 +

                       x> 0 then

               a(x+)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

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

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

b)      

Solution

Table of values of y= x²-4x+2

Find the maximum value of the function.;

Maximum value

         =  

Y= (     ) = – 2

Find 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

   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

f(x) =5-x-9x²

Solution

  f(x) = -9x² – x + 5

      =   -9( x²  –  )   +  5

      = -9(  x² –  +  )   +  5 +  

    = -9(x –  )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

Solution

Maximum value          )         

  )  

The turning point of the graph is     ) 

The minimum value of the graph is y =  – 
axis of symmetry = 
POLYNOMIAL FUNCTIONS 

The polynomial functions are the functions  of the form, ” P(x) =a_n xⁿ + a _n-1 x^n-1  + a_n-2 x^n-2  + . . . +  a₁ x¹+a₀ x^{0 “.} Where n is non negative integer and a_n, a_n-1, a_{n-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

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}

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

  Domain

   = {x :  XR}

  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?

  

(a) Domain of f(x )  = {   x: x    }

(b)  •Range  of f(x)  = {    y : y }
•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 |

 

 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) = n^x where n is called base and x is called exponent.

  Example
1. Draw the graph of f(x)=2^x

      Solution :

        Table of values

                     
     

2. Draw the graph of f (x)=2^-x

Solution:

Table of values

 

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.

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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

Example

Draw the graph of f(x)=

Solution

Table of values

 

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) = 2^x and f(x)= under the same graph.

Solution

(i) y= 2^x

Apply log on both sides
log x=log 2^{y
log x  =y log 2
y = log (x- 2)
f-1(x) = log (x- 2)}
(ii) f(x)=
y=

x=

y= 2^x

Table of values

 

2. Draw the graph of the function f(x)=3^x

Table of values

3. Draw the graph of the function f(X)=8^x

Solution

Table of values

                       

Exercise

1. Find the graph of y =2^x 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

Copy and complete the following table and hence draw the graph of f(x)=(1/4)^x

OPERATION OF POLYNOMIAL FUNCTION

ADDITIONAL AND SUBTRACTION

The sum of two polynomials is found by adding the coefficients of terms of the same degree or like terms, and subtraction can be found by subtracting the coefficient of like terms.

EXAMPLE

If p(x) = 4x²– 3x+7, Q(x) =3x+2 and r(x)= 5x³-7x²+9

Solution

Sum =p(x)+q(x)+r(x)

(4x²-3x+7)+(3x+2)+(5x³-7x²+9)
4x²-3x+7+3x+2+5x³-7+9

4x²+5x²+7x³+3x+7+2+9

5x³-3x²+18

Alternatively the sum can be obtained by arranging them vertically

              4x²-3x+7

               0x²+3x+2

        5x³-7x²+0+9

        5x³-3x²+0+18

P(x)+q(x)+r(x)=5x³-3x²+18

Subtract -5x²+9+5 from 3x²+7x-2

Solution

(3x²+7x-2)-(-5x²+9x+5)

3x²+7x-2+5x²-9x-5

3x²+5x²+7x-9x-2-5

8x²-2x-7 answer

MULTIPLICATION
The polynomial R(x) and S(x) can be multiplied by forming all the product of terms from R(x) and terms from S(x) and then summing all the products by collecting the like terms.
The product can be denoted by RS(X)

example

If p(x)=2x²-x+3 and q(x)=3x³-x find the product p(x)q(x)

Solution

P(x) q(x)= (2x²-x+3) (3x³-x)

6x⁵-2x³-3x⁴+x²+9x³-3x

6x⁵-3x⁴+7x³+x²-3x

DIVISION
The method used in dividing one polynomial by another polynomial of equal or lower degree is the same to the one used for the long division of number

Example:
1. Given p[x] =x³-3x²+4x+2 and q[x]=x-1 find

Solution

   

                              x³-x²

                              0-2x²+4x

                                -2x²+2x

                                  0+2x+2

                                       2x-2

                                       0+4

  Therefore

   =     x² – 2x + 2 +

This means

X³– 3x²+4x+2 is dividend x-1 is divisor

X²-2x+2 is the quotient and 4 is the remainder

Then

X³-3x²+4x+2        =       (x²-2x+2) (x-1)       +         4

Divided                        quotient   x divisor + remainder

2. Divide p(x) = x³-8 by q(x) = x-2

Solution:

                 

                                          -x³ -2x²

                                                 2x²-8

                                                -2x²-4x

                                                        4x-8

                                                         4x-8
                                                         –   –

In the example [2] above there is a remainder i.e. the remainder is zero so dividing one function by another function is one of the way of finding the factors of the polynomial thus if we divide p(x) by q(x) is one of the factors of p(x) other factors can be obtained by fractionazing the quotient q(x).

3. Given  p(x) = x³-7x+6 and q(x) = x+3 determine whether or not d(x) is one of the factors of p(x) and hence find the factors if p(x)

Solution

                      

                                                X³+3x²

                                                    -3x²-7x

                                                     -3x²-9x

                                                             2x+6

                                                             -2x+6

                                                              –     –

Since the remainder is zero d(x) =x+3 is one of the factors of p(x)

To factorize

  X²-3x+2

(x²-x) – (2x+2)

x(x-1)-2(x-1)

(x-2)  (x-1)

Therefore other factors are (x-2) and (x-1)

THE REMAINDER THEOREM

The remainder theorem is the method of finding the remainder without using long division

Example

1. If p(x) = (x-2)   and  q(x)+8. Dividend divisor quotient remainder, Then by taking  x-2=0 we find

x-2= 0
x=2

subtracting x=2 in p(x)

p(2) =2-2 q(x) +r

p(2)= 0 x q(x)+r

p(2)= 0+r

p(2) =r

So the remainder r is the value of the polynomial p(x) when x=2

2. Give p(x) = x³-3x²+6x+5 is divided by d(x) = x-2 find the remainder using the remainder theorem

Solution

Let d(x)=0

x-2=0

X-2+2 =0+2

X=2

Subtracting x=2 in p(x)

P(2)= 23-3[2]2+6[2]+5

P(2)=8-12+12+5

P(2)=8+5

P(2)=13

The remainder is 13

3. The remainder theorem states that if the polynomial p[x] is divided by [x-a] then the remainder ‘r’ is given by p[a]

P(x) =(x-a) q(x)+r hence

P(a) =(a-a) q(x)+r

P(a)=o(q(x))+r

P(a)=0+r

P(a)=r

More Examples

1. By using the remainder theorem, Find the remainder when p[x] =4x²-6x+5 is divided by d[x] =2x-1

Solution

    d (x)=0

    2x-1=0

2x-1+1=0+1

X =  

Substituting

X=      in p(x)

P() =4  x -6()+5

P()=1 – 6 x  +5

P()=1 -3+5

P()=  3

The remainder is 3

2. P(x)= 3x²-5x+5 is divided by d(x) =x+4

Solution

Let d(x)=0

X+4=0

X+4=0-4

X=-4

Subtracting x=-4 in p(x)

P(-4)=3(-4)²-=5(-4)+5

P(-4)=48+20+5

P(-4)=68+5

P(-4)= 73

The remainder is 73

3. P(x)= x³+2x²-x+4 is divided by d(x) = x+3

Solution

Let d(x)=0

X+3=0-3

X=-3

Substituting x=-3 in p(x)

P(-3)= -3³+2(-3)2-(-3)+4

P(-3)=-27+18+3+4

P(-3)=-27+21+4

P(-3)=-27+25

P(-3)=-2

The remainder is -2

4. Find the value of ‘a’ if x³-3x²+ax+5 has the remainder of 17 when divided by x-3

Solution

By using remainder theorem

Let x-3 =0

x-3=0+3

x=3

Substituting x=3 we have

x³-3x²+ax+5=17

(3)³-3(3)²+a(3)+5=17

27-27+3a+5=17

3a+5=17-5

3a=12

a=4

5. If ax²+3x-5 has a remainder -3 when divided by x-2. Find the value of a.

Solution

 By using the remainder theorem

Let x-2=0

x-2=0+2

x=2

subtracting x=2 we have

a(2)2+3(2)-5 =-3

4a+6-5= -3

4a+1= -3

4a= -4

a = -1

Exercise

1. Divide p(x) by d(x) in the following

P(x)=2x²+3x+7 d(x)=x²+4

Solution

                       

Use remainder theorem to find the remainder when;

1. P(x)=x³-2x²+5x-4 is divide by d(x)=x-2

Solution

Let d[x]=0

x-2 + 2=0+2

x=2

substituting x=2 in p(x)

p(2)=2³-2[2]²+5[2]-4

p(2)=8-8+10-4

p(2)=10-4

p(2)=10-4

p(2)=10-4

Therefore the remainder is 6

2. p(x)= 2x⁴+ x³+x-  is divided by d(x)=x+2

solution

let d(x)=0

x+2 – 2 = 0-2

x = -2

Substituting x = – 2 in p(x)

p(-2)=2(-2)⁴+(-2)³+(-2)( –)

p(-2)= 2(16)+(-8)-2( –)

p(-2)=32-8-2 ( –)

p(-2)=24-2( –)

p(-2)=21

Therefore the remainder is 21