Halloween 2015

Long and Synthetic Division

Hi there! Let's review the first part that we've covered in Unit 3 so far. In this unit, we mostly use division to find the answers, which could be either Long Division or Synthetic Division. However, there are certain points that we have to remember when use these methods.


Example 1: Divide 16t + 6t³ - 4- 19x²  by t -2

Long division                                                                     

-Always arrange the terms from                                      

highest to lowest degree

-Use subtraction                                                               

                                                                                             

OR WE CAN USE,

Synthetic Division

 -The arrangement of the coefficients is based from

 highest to lowest degree

 -Use Addition

-The divisor will be 2 because (x-a) → t-2 → x=2

• 6t³ -19t² +16t -4 ÷  t -2 =  6t² -7t +2
• Factor the quotient 6t² -7t +2 

       6t² -7t +2                                                            

       6t² -4t -3t +2

       2t (3t-2) -1 (3t -2)

       (2t -1) (3t-2)

Answer:  6t³ -19t² +16t -4 =  (2t -1) (3t-2) (t -2)

A remainder of 0 means the divisor is a factor of the given polynomial

Example 2: Divide x -2 +4x² +x³ by x -1

Long Division                                                                                        

-Arrange the terms from                                                                    

highest to lowest degree again                                                         

-Use subtraction

OR WE CAN USE,

 Synthetic Division

 -The arrangement of the coefficients is based from

 highest to lowest

  -Use Addition

-The divisor will be -3 because (x-a) → t+3 → x= -3

x³ +4x² +x -2 ÷  x -1 =  x² +5x +6

Factor the quotient x² +5x +6

x² +5x +6

x² +3x +2x +6

x(x+3) 2(x+3)

(x+2) (x+3)

A remainder of more than 0 means the divisor is not a factor of the given polynomial

To conclude, it's way easier to use synthetic division but try doing long division once in a while since it's a good practice. 

Going Backwards with Transformations

       A couple of days ago, we learned how to graph y= f(x) if the transformed function is given. We can determine the original function y=f(x) when the transformed function is given, by first determining what transformation(s) occurred in the function. Then, reverses the steps of transformation(s) after doing so, apply the opposite operation to the x and y point(s).

Example:

-        Given the transformed function y= 2f ½ (x+2) -1 below. What is the graph of y= f(x).

       Transformation:             To get y= f(x)                        

     - Multiply y-values by 2            - Add 1 from y-values

     - Subtract 1 from y-values         - Divide y-values by 2

     - Multiply x-values by 2           - Add 2 from x-values

     - Subtract 2 from x-values        - Divide x-values by 2

                        

     

                                                        y= 2f ½ (x+2) -1        y= f(x)

                                          (-1, 1)                      (½, 3/2)

                                                          (2, -2)                      (5/2, ­­-½)

                                                          (4,-2)                       (4, -½)

                                                          (6, 0)                       (5, ½)

Hello friends
So far we have been learning about function transformations in class, and about inverse of a function. 
This post is about how to find inverse of a function algebraically. Let us consider an example.

Example:Let f:R→R    be defined by f(x)=3x+1  . Find the inverse function f −1   .


Solution: For any input x  , the function machine corresponding to f  spits out the value y=f(x)=3x+1  . We want to find the function f −1   that takes the value y  as an input and spits out x  as the output. In other words, y=f(x)  gives y  as a function of x  , and we want to find x=f −1 (y)  that will give us x  as a function of y  .
To calculate x  as a function of y  , we just take the expression y=3x+1  for y  as a function of x  and solve for x  .

yy−1y−13  =3x+1=3x=x  

Therefore, we found out that x=y/3−1/3  , so we can write the inverse function as

f −1 (y)=y3 −13 . 

In the definition of the function f −1   , there's nothing special about using the variable y  . We could use any other variable, and write the answer as f −1 (x)=x/3−1/3  or f −1 (★)=★/3−1/3  . The placeholder variable used in the formula for a function doesn't matter.
To verify that f −1   is really the inverse of f  , we should show that the composition of f  and f −1   doesn't do anything to the input. In this case, the order shouldn't matter, and the functions f∘f −1   and f −1 ∘  f$ should both do nothing. Let's check this.
First, we apply f  followed by f −1   .

(f −1 ∘f)(x) =f −1 (f(x))=f −1 (3x+1)=(3x+1)/3−1/3=x+1/3−1/3=x  

Second, we apply f −1   followed by f  .

(f∘f −1 )(x) =f(f −1 (x))=f(x/3−1/3)=3(x/3−1/3)+1=x−1+1=x  

In both cases, applying both f  and f −1   to x  gave us back x  . Indeed, f −1 (x)=x/3−1/3  .