NAMA : RANI RAHAYU P.
NIM : 07305144001
PRODI : MATEMATIKA NR’07
PROGRAM : BHS. INGGRIS II
EMAIL : ranny_15sept@yahoo.com
ranny.mathuny@gmail.com
CP : 08562544575


I. Pre-Calculus : Graph of a rational function.
Graph of a rational functionàcan have discontinues has a polynomial in the denominator.
Example :
f(x) = (x+2) / (x-1), if we insert x = 1, we get
f(1) = (1+2) / (1-1)
= 3 / 0àbecause the denominator is 0 is not possible, so we don’t insert x = 1, so x = 1 is bad choice.
If we insert x = 0, we get
f(0) = (0+2) / (0-1)
= 2 / (-1)
= -2 àwe can write (0,-2), so x = 0 is not bad choice.
Rational function denominator can be zero, not all rational function will give the not possible solution. Polynomialàx zero in the denominator (break in the graph, missing point, feasible, and not allowed).

Example :
y = (x2-x-6) / (x-3) if we insert x=3
= [(3)2-3-6] / (3-3)
= 0 / 0àis not possible.
y = (x2-x-6) / (x-3), we can write
= (x-3) (x+2) / (x-3), the numerator get factor (x-3) and (x-2) we can delete with the denumerator (x-3),so the equation become
y = x+2, if we insert x = 3 we can get
y = 5
(3,5)
The conclusion is if we insert x = 3 and the numerator can not factor we get not possible solution, but if we get the factor from the numerator we can get the posibble solution.

II.Limit by Inspection
Limit by inspection there is two condition
1. positive and negative invinitif
2. ………………………………
Example :
§ lim (x3+x4) / (x2+x+1) = ¥
xà¥
· polynomial over polynomial
· x approaches invinity
à if degree of the numerator f(x) > degree of the denumerator g(x) and the highest power of the numerator f(x) is positive, so :
lim f(x) / g(x) = ¥
xà¥
à if degree of the numerator f(x) > degree of the denumerator g(x) and the highest power of the numerator f(x) I negative, so :
lim f(x) / g(x) = -¥
xà¥
§ lim (x2+3) / (x3+1) = 0
xà¥
à if the degree of the numerator f(x) < degree of the denunerator g(x), so :
lim f(x) / g(x) = 0
xà¥
§ lim (4x3+x2+1) / (3x3+4) = 4 / 3
xà¥
à if the degree of the numeratoe f(x) = degree of the denumerator g(x), so :
lim f(x) / g(x) = coefficient the highest power from f(x) / coefficient the higgest xॠpower from g(x)


III. EXERCISE
1. The figure above shows the graph of y=g(x) in the function h is definicied by h(x)=g(2x)+2, what is the value is h(1)…..
h(x)=g(2x)+2,insert x=1
=g(2.1)+1
=g(2)+2,because g(2)=1 (from the graph), so we can write
=1+2
h(x) = 3
à so the value h(1)=3
2. Let the function f be definiced b f(x)=x+1 if 2f(p)=20, what is the value of f(3p)..
f(3p) àwhat is f when x=3p
f(x)=x+1
2f(p)=20
f(p)=10
f(x)=x+1,insert x=p
f(p)=p+1, because f(p)=10 so,
10=p+1
p=9
f(x)=x+1
f(3p)=3p+1, insert x=3p
because p=9, so we insert at f(3p)=3p+1, we get
f(3p)=3.9+1
=27+1
=28
3. In the xy coordinate plane, the graph of x=y2-4 intersecte line that (0,p) and (5,t). What is the greatest possible value of the slope f….
Greatest m
Curve x=y2-4 through (0,p) and (5,t) we get line g through (0,p) and (5,t) the slope from line g is (y2-y1) / (x2-x1) = (t-p) /(5-0), so the slope from line g is (t-p) / 5.
So the gretest value possible value of the slope f is m= (t-p) / 5

IV.Inverse Function
y=f(x)
x=g(y) àinvertible
Example :
f(x)=2x-1
g(x)=1/2x+1/2
f(g(x))=2(1/2x+1/2)-1, insert x=g(x)
g(f(x))=1/2(2x-1)+1/2, insert x=f(x)
àf(g(x))=gof composition of function
àg(f(x))=fog
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), inverse from this function is
y=(-1-2x) / (x-1)
from the two function above we can find the value from x and y
Example: cx=0, so 0=y-1
y=1
cy=0, so 0=-1-2x
x= -1/2
2x-1=y
2x=y+1
x=1/2 (y+1)
x=1/2y+1/2, inverse from this function is
y=1/2x+1/2