Mathematics 1 summary midterm - UNIVERSITY OF AMSTERDAM

Mathematics 1 – midterm

Pre-knowledge

ℕ Natural numbers 1, 2, 3, 4, …
ℤ Integers …, -3 , -2, -1, 0, 1, 2, 3, …
ℚ Rational numbers Any number in the form of p/q where p and q are integers and q is not equal to 0
is a rational number. Examples of rational numbers are 1/2, -3/4, 0.3, or 3/10.
ℝ Real numbers Rational and Irrational numbers; Irrational numbers cannot be written as a
fraction (examples: e , π , √ 2 )

A2=0 ⟺ A=0
A ∙ B=0 ⟺ A=0 ∨ B=0
A ∙ B> 0 ⟺(A >0 ∧ B>0) ∨( A <0 ∧ B<0)
A ∙ B< 0 ⟺(A >0 ∧ B<0) ∨( A <0 ∧ B>0)

Bounded, closed: [ a , b ]={x ϵ R∨a ≤ x ≤b }
Bounded, open: (a , b)={x ϵ R∨a< x <b } [0,5]  x=0 en x=5 doen wel mee
Unbounded, open: (a , ∞) or (−∞, b)
Bounded, half open, half closed: ¿ or ¿ (0,5)  x=0 en x=5 doen niet mee
Unbounded, half open, half closed: ¿ or ¿

Functions of one variable
A function of a real variable x with domain Df is a rule that assigns a unique (vertical line test) real number to
each number x in Df. As x varies over the whole domain, the set of all values f (x) is called range of f , Rf.
f Symbol for the function
f ( x ) Value of f at x
x Independent (exogenous) variable or argument y=f (x )
y Dependent (endogenous) variable

Finding the domain of a function
If f ( x )= √ g( x ) then g( x )≥ 0 Numerator
t (x) Denomerator
If f ( x )= then n( x )≠ 0
n( x)
If f ( x )=log ( g ( x )) then g( x )> 0

Linear functions (polynomials of degree 1)
f ( x )=ax+ b, a ≠ 0
a is the slope and b is the y-intercept


a> 0: the graph of f is an increasing line
a=0: the graph of f is a horizontal line
a< 0: the graph of f is a decreasing line

Obtaining the equation of the straight line through (x 1 , y 1 ) and (x ¿ ¿ 2 , y 2 )¿ .
y 2− y 1 Δ y
1. Compute the slope a= =
x 2−x 1 Δ x
Implicit function: y− y1 =. ..
Explicit function: y=.. .

, 2. y− y1 =a( x−x 1 ); substitute a
Quadratic functions (polynomial of degree 2)
f ( x )=a x2 +bx +c , a ≠ 0
The graph of a quadratic function is a parabola.

−b
Line of symmetry: x=
2a
−b
If a> 0 then f ( x )=a x2 +bx +c has a minimum at x=
2a
−b
If a< 0 then f ( x )=a x2 +bx +c has a maximum at x=
2a

Three methods of finding intersection point(s) with x -axis ( y=0):
I. Use quadratic formula (ABC formule)
II. Factorize the quadratic function (not always possible)
III. Completing the squares (kwadraat afsplitsen)

Method of using the quadratic formula
2
f ( x )=a x +bx +c , a ≠ 0
−b ± √ b −4 ac
2
x 1,2=
2a
2
D=b −4 ac is called the discriminant.

Method of factorizing the quadratic function
Idea: use that ( x +a ) ( x+ b )=x 2 +ax +bx +ab=x 2+ ( a+b ) x +ab

Method of completing the squares
Idea: write the quadratic formula as a product of two linear functions and Here you can directly see
‘a remainder’. where the
General formula: f ( x )=a x2 +bx +c=a ¿ minimum/maximum is at

Polynomials of degree n
Cubic functions (polynomials of degree 3)
f ( x )=ax 3+ b x2 +cx + d , a ≠ 0

Polynomials of degree n:
n n−1
P ( x )=an x + an−1 x +. . .+a1 x+ a0, a ≠ 0, a ’s are constants (coefficients)
D P(x) =R

If P ( a ) =0 then P ( x )=(x−a)P1 ( x ) with P1 ( x) a polynomial of degree n−1
The polynomial P1 ( x) can be found by long division (staartdelingen).

Rational functions:
n n−1
P (x) an x +a n−1 x + .. .+ a1 x +a 0
f ( x )= = (Q(x )≠ 0)
Q(x ) b m x m +b m−1 x m−1 +. ..+ b1 x +b 0
With P( x ) and Q(x ) polynomials.
Domain f ( x ) := { x ϵ R|Q( x)≠ 0}
f ( x )=0 ⟺ P ( x )=0 and Q( x )≠ 0
f (x) is called proper if Degree ( P ( x ) ) < Degree (Q ( x ))