tugas 1

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:

1. 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)

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 :

1. X goes to (+) or (-) infinity

2. limit involves a polynomial divided with polynomial

for example :

lim 2x³+6 / x²+2x+1 ,is approaches infinity

x ~

this problem have 2 condition :

1. polynomial over polynomial

2. 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 :

• Must be dividing by polynomials

• X has to be approaching infinity

Shortcut rules :

1. 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

2. 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

3. 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

• continuity

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)

1. 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

• X = 0

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 = log2^x

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 :

x	y
0	p
5	t