Pre-Calculus
Graph of a rasional function
A. Rasional number is number able to be expressed a/b which b=0
For example (discontinuity)
a) f(x) =x+4/x-1
If substitution of x=1 so f(1) = 1+4/1-1
= f(1) =5/0
f(x) =x+4/x-1 imajiner when x=1
If substitution of x= -4so f(-4) = (-4)+4/-4-1
= f(-4) =0/-5
The co-ordinate is (-5,0)
f(x) =x+4/x-1 having value If x≠1,
· f(x) =x+4/x-1 discontinuity
Hp:{x x <> 1, x R} or can we white(-∞,1)and (1,∞)
Example ( continuity )
b) g(x) =1/x2+1
The function of g(x) always having value because the dominator don’t consist of zero for x R.
If x = 1, g(1) =1/2
x = 2, g(2) = 1/5
Graph g(x) = x/x2+1
· g(x) = 1/x2+1 continuity
Hp={x x R} or can we write (∞,1)
Polynomial
There are two ways to get the value from polynomial function
1. Subtitute of number for the variable.
Example:
g(x) =x2+x-6/x-3
x=3, baaad! (not allowed)
Ø This method used direct by including value x
Ø g(x) discontinuity if x=3 because g(x) = 0/0
2. Factor of the polynomial function
Example:
g(x) = x2+x-6/x-3
We can formulate x2+x-6to(x-3)(x+2),so
g(x) =(x-3)(x+2)/x-3
= x+2/1
g(x) = x + 2
Determining Limits
There are 2 conditions to determining limits by inspection
1. x goes to positive or negative infinity
2. limit involves a polynomial divided by a polynomial
For example:
limx→∞x4- 4/x3+x+1
This problem on two conditions:
1. polynomial over polynomial
2. x approaches infinity
To apply these rules:
· Must be dividing by polynomials
· X has to be approaching infinity
If the highest power of x is greater in numerator so the limit is positive or negative infinity
Example:
1.
- limx→∞ x4- 4/x3+x+1
- Highest power of x in numerator is 4
- Highest power of x in denominator is 3
Since all the number are positive and x going to positive infinity so value the limit is infinity.
- Limx→∞ x4- 4/x3+x+1
If you can’t tell if the answer is positive or negative infinity:
Ø You can substitute a large number for x
Ø See if you end up with a positive or negative number
Ø Whatever sign you get is the sign of infinity for the limit
2. Limx→∞ x2-2/x3+4x+5
Used when:
Ø Highest power of x in numerator is SAME as highest power of x in denominator
Ø Limit x going to (positive or negative)
The quotient of the coefficients of two highest power
Limx→∞ x2-2/x2+4x+5 = 1/1= 1
Inverse Function
F(x,y)=0
Function y = f(x) : Vertical Line
1.1. Function x = g(y) : Horizontal Line
For example:
1. y = x2
x=g(y) : Horizontal Line : Invertible
2. y = 2x -1
Graph:
x
Y
0
-1
1/2
0
y = x other line on the picture
- substitute y = x to y = 2x-1
x =2x-1 making a move internodes (from right to left)
≈ x-2x = -1
≈ - x = -1 multiple with -1
≈ x = 1…………………………….( i )
Substitute ( i ) to y = x (or y = 2x – 1)
y = 1
so intersection two line y = x and y = 2x – 1 on (1,1)
From y = 2x -1, we will change to become function of x, so
y = 2x -1 making a move internodes (from right to left)
≈ y – 1 = 2x multiple with 1/2
≈1/2y+1/2=x
≈1/2(y+1)=x
Because y = x and x =1/2(y+1)
y = 1/2 (x+1)
We find new function!
y = ½(x+1)
from all that we calculate above, we get:
f(x) = 2x – 1
g(x) = ½(x+1)
f( g(x) ) we substitute g(x) to variable on the function f.
g( f(x) ) we substitute f(x) to variable on the function g.
Ø f ( g(x) ) = 2 ( g(x) ) - 1
f ( g(x) ) = 2(1/2(x+1)-1
f ( g(x) ) = x + 1 – 1
f ( g(x) ) = x
Ø g ( f(x) ) = ½((2x-1)+1
g ( f(x) ) = ½((2x))
g ( f(x) ) = x
g =f-1
f ( g(x) ) = f (f-1(x)) = x
g( f(x) ) = f (f-1(x)) = x
When y = 0 -1 – 2x = 0
≈-2x = 1
≈ x = -1/2
Graph of a rasional function
A. Rasional number is number able to be expressed a/b which b=0
For example (discontinuity)
a) f(x) =x+4/x-1
If substitution of x=1 so f(1) = 1+4/1-1
= f(1) =5/0
f(x) =x+4/x-1 imajiner when x=1
If substitution of x= -4so f(-4) = (-4)+4/-4-1
= f(-4) =0/-5
The co-ordinate is (-5,0)
f(x) =x+4/x-1 having value If x≠1,
· f(x) =x+4/x-1 discontinuity
Hp:{x x <> 1, x R} or can we white(-∞,1)and (1,∞)
Example ( continuity )
b) g(x) =1/x2+1
The function of g(x) always having value because the dominator don’t consist of zero for x R.
If x = 1, g(1) =1/2
x = 2, g(2) = 1/5
Graph g(x) = x/x2+1
· g(x) = 1/x2+1 continuity
Hp={x x R} or can we write (∞,1)
Polynomial
There are two ways to get the value from polynomial function
1. Subtitute of number for the variable.
Example:
g(x) =x2+x-6/x-3
x=3, baaad! (not allowed)
Ø This method used direct by including value x
Ø g(x) discontinuity if x=3 because g(x) = 0/0
2. Factor of the polynomial function
Example:
g(x) = x2+x-6/x-3
We can formulate x2+x-6to(x-3)(x+2),so
g(x) =(x-3)(x+2)/x-3
= x+2/1
g(x) = x + 2
Determining Limits
There are 2 conditions to determining limits by inspection
1. x goes to positive or negative infinity
2. limit involves a polynomial divided by a polynomial
For example:
limx→∞x4- 4/x3+x+1
This problem on two conditions:
1. polynomial over polynomial
2. x approaches infinity
To apply these rules:
· Must be dividing by polynomials
· X has to be approaching infinity
If the highest power of x is greater in numerator so the limit is positive or negative infinity
Example:
1.
- limx→∞ x4- 4/x3+x+1
- Highest power of x in numerator is 4
- Highest power of x in denominator is 3
Since all the number are positive and x going to positive infinity so value the limit is infinity.
- Limx→∞ x4- 4/x3+x+1
If you can’t tell if the answer is positive or negative infinity:
Ø You can substitute a large number for x
Ø See if you end up with a positive or negative number
Ø Whatever sign you get is the sign of infinity for the limit
2. Limx→∞ x2-2/x3+4x+5
Used when:
Ø Highest power of x in numerator is SAME as highest power of x in denominator
Ø Limit x going to (positive or negative)
The quotient of the coefficients of two highest power
Limx→∞ x2-2/x2+4x+5 = 1/1= 1
Inverse Function
F(x,y)=0
Function y = f(x) : Vertical Line
1.1. Function x = g(y) : Horizontal Line
For example:
1. y = x2
x=g(y) : Horizontal Line : Invertible
2. y = 2x -1
Graph:
x
Y
0
-1
1/2
0
y = x other line on the picture
- substitute y = x to y = 2x-1
x =2x-1 making a move internodes (from right to left)
≈ x-2x = -1
≈ - x = -1 multiple with -1
≈ x = 1…………………………….( i )
Substitute ( i ) to y = x (or y = 2x – 1)
y = 1
so intersection two line y = x and y = 2x – 1 on (1,1)
From y = 2x -1, we will change to become function of x, so
y = 2x -1 making a move internodes (from right to left)
≈ y – 1 = 2x multiple with 1/2
≈1/2y+1/2=x
≈1/2(y+1)=x
Because y = x and x =1/2(y+1)
y = 1/2 (x+1)
We find new function!
y = ½(x+1)
from all that we calculate above, we get:
f(x) = 2x – 1
g(x) = ½(x+1)
f( g(x) ) we substitute g(x) to variable on the function f.
g( f(x) ) we substitute f(x) to variable on the function g.
Ø f ( g(x) ) = 2 ( g(x) ) - 1
f ( g(x) ) = 2(1/2(x+1)-1
f ( g(x) ) = x + 1 – 1
f ( g(x) ) = x
Ø g ( f(x) ) = ½((2x-1)+1
g ( f(x) ) = ½((2x))
g ( f(x) ) = x
g =f-1
f ( g(x) ) = f (f-1(x)) = x
g( f(x) ) = f (f-1(x)) = x
When y = 0 -1 – 2x = 0
≈-2x = 1
≈ x = -1/2
Function & Graphs
13. The figure above shows the graph of y = g(x).
If the function his defined by h(x) = g(2x) + 2. What is the value of h(1)?
Solution:
h(x) = g(2x) + 2
because we need get the value of h(1), so we can substitute 1 for x
h(1) = g(2*1) + 2
h(1) = g(2) + 2
is absis.
Ø g(2) is ordinate when the absis=2.
Ø (2 , g(2)) on h(x)= g(2x) +2
We find ordinate when x=2 which (2,y) on h(x)
n From the picture we can see y=1
Ø (2, g(2)) = (2,y)
g(2) = y
g(2) = 1
h(1) = g(2) +2
h(1) = 1 + 2
h(1) = 3
17. In the xy –coordinate plane, the graph of x = y2-4 intersects line A at (0,p) and (5,t). What is the greatest possible value of the slope of A?
Solution:
greatest m
x = y2 - 4
x y
0 p
5 t
