Nur rahmani ristiamarta MAT NR07
(07305144011)
english 2
VIDEO 1
PRE CALCULUS
Graph of a rational function which can have discontinuiteis has a polynomial in the denominator.
For example :
F(x) = (x+2) / (x-1) , x = 1
F (1) = (1+2) / (1-1) = 3/0 when x = 1,
is a bad idea because of 0 in the denominator and it’s say break function graph.
F (x) = (x+2) / (x-1)
Insert x = 0 f (0) = 2/-1 = -2
Insert x = 1 f (1) = 3/0 (it’s impossible)
discontinuity
break
x=1
Rational Functions
Now rational functions don’t always work this way. Not all rational functions will give zero (0) in denomirator. Don’t forget rational functions denominator can be zero (0)
F (x)= 1/(x2+1)
If x= -1 f (-1)= 1/(-12+1)
Will never zero because of + 1
For polynomial the graph is a smooth un broken curved. For rational functions,
x zero in the denuminator that is an impossible situation. So there is no value for the function and it’s break in the graph.
Break is in two ways:
Break missing point
For example :
y= x2-x-6 / x-3
y(3)= 32-3-6 / 3-3
= 0/0 this is typical example of missing point syndrome
it’s not possible, not feasible, not allowed
x=3
when you see the result of 0/0, it’s also tells you that should be possible that the factor top and bottom rational function
and simplify.
Example:
y= x2-x-6 / x-3
y= (x-3)(x+2) / (x-3), (x-3)in the top canceled
to (x-3) in the bottom
so the function y= (x+2)
Missing point is loophole
for the original function y= x2-x-6 / x-3 without simplify x=3 is a bad point. So we must simplify first there is no problem for x=3
REMOVABLE SINGULARTY
Appears simply a missing point in the graph, when “x” leads to 0/0 for this kind of break if a factor and simplify, rational funtions division by zero can be avoided.
VIDEO 2
LIMIT BY INSPECTION
Limit by inspection there are 2 condition :
X goes to (+) or (-) infinity
limit involves a polynomial divided with polynomial
for example :
lim 2x3+6 / x2+2x+1 ,is approaches infinity
x ~
this problem have 2 condition :
polynomial over polynomial
X approaches infinity
If the problem lim 2x3+6 / x2+2x+1 using time procedures.
x ~
the key to determining limits by inspection is in looking at the power of x in the numerator and the denominator.
Remember that to apply these rules :
Shortcut rules :
if the highest power of x is the greater in numerator than the denominator, and the limit is possitive or negative infinity
example : lim 2x3 + 6 / x2+2x+1 , the limit of this expression is a + or – infinity
x ~
since all the numbers are + and x going to + infinity, the limit must be + infinity.
The highest power of x in numerator is greater than in the denominator, so the value is ~ .
If you can’t tell if the answer is + or -, you can :
substitute a large number of x
see if you end up with a possitive or negative number
whatever sign you get is the sign of infinity for the limit
if the highest power is in the denominator, the limit is 0
example : lim (x3 + 2) / (x4+8) , since the highest power of x is in numerator = 3, and x ~ the denominator =4,the value is 0
if the highest power of x on numerator is same as the highest power in denominator, limit is the coefficient of the highest power of x in numerator and denominator.
If this is a key in the limit is just the quotient of coefficients of the two highest powers. Remember the coefficient : the number that goes with a variable.
According to this rule : that means lim = coefficients of x3’s over each other.
example : lim (5x4+6x3-2x+1) / (10x4-2) , the highest power of x is same as in
x ~ numerator and denominator, the value of the limit is the coefficient = 5/10
normally we with say something continuous if it has no breaks or disruptions.
VIDEO 4
INVERSE FUNCTION
Example 1:
F(x,y) = 0
Function y=f(x) = VLT (Vertical Line Text)
function if also x=g(x) = HLT (Horizontal Line Text) Invertibel
example curve x2
0<x
Y=x
Y= 1/2x + 1/2
(0, 1/2)
(0,-1/2)
(0,1)
(0,-1)
Y = 2x – 1
Y = x
X = (2x-1)
X=1
2x-1=y
2x = y+1
X = ½(y+1)
X = ½y + ½
Y = ½x + ½
F(x) = 2x – 1
G(x) = ½ x + ½
F(g(x))= 2 [ (½ x + ½) ] – 1
= x + 1 – 1 = x
G(f(x))= ½ [(2x – 1)] + ½
= x - ½ + ½ = x
G=f-1
F(g(x))= f (f-1(x))
= x
G(f(x))= f-1(f(x))
Example 2:
Y= x-1 / x+2
X= 1
(1,0)
(0,-1)
X= -2
Y= x-1 / x+2
Y (x+2) = (x-1)
Yx + 2y = x-1
Yx – x = -1 – 2y
(y-1) x = -1 – 2y
X= -1-2y / y-1
Y= -1-2y / x-1
Y = -1
* y = o
-1 – 2x = 0
-2x = 1
X = -1/2
V asymtot X = 1
H A @ Y = -2
X= -2
Y = 2 x
Y = log2x
Video 3
English Solving Problem Graph Math
13.) The Graphs of y= g(x). If the h is defined by h(x) = g(2x)+2. What is the value of h(1)?
Answer :
h(1) h(x) = g(2x)+2
h(1) = g(2)+2
= 3
13.) Let the function f defined by f(x) = x + 1 if 2f(p) = 20, what is the value of f(3p)?
Answer :
F(3p), what is f when x = 3p?
, So,
17.) In the xy. Coordinate plane, the graph of , intersect line l at (0,p) and (5,t). What is the greatest possible value of te slope of l?
Answer :
Greatest m?
line l :
