Summary Mathematics for Pre-Master Mid-Term TISEM Tilburg University
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Course
350897-B-6 (350897B6)
Institution
Tilburg University (UVT)
This is an extensive summary of the video lectures of each week (class notes) screens of important example questions discussed during the Q&A, and more examples. Studying this and practicing the old exams will guarantee that you pass the course! Good luck.
Content Exam October:
From book Mathematics for Business Economics by Hamers, Kaper and Kleppe:
- Chapter 1
- Chapter 2, except 2.3.3
- Chapter 3
- Chapter 4.1 and 4.2
- Chapter 5.1, 5.3 and 5.7
- All examples related to minimum function are dismissed.
Lecture Week 1, Chapter 1
Function
= With a function you have one or more input variables (independent variable), and an output
variable (dependent variable).
Example: input variable = x, and then the function is x2 (whatever you put in, you must square it), and
the output is y.
So the function is: y(x) = x2
D(p) = 10-p
z(xy) = √x + xy2
Domain (D) of a function = set of values that you can use as an input.
Depends on what the input variable is, example: price (p) cannot be negative so p ≥ 0. Also depends
on what the output variable is, example: demand (D) cannot be negative.
So, for D(p) = 10-p → 0 ≤ p ≤ 10 OR p∈[0,10]
if boundaries are included: .. ∈[.. , ..]
if boundaries are not included: .. ∈(.. , ..)
x∈(0,∞)
Another example: z(xy) = √x + xy2 → x cannot be negative, as it is put in a square root in the function.
So x ≥ 0.
Range = set of all possible outputs.
Example:
1
,Input: litres of petrol sold (q)
Output: revenue (R)
Function: R(q) = 1.65q
Domain: 0 ≤ q ≤ 5000
Range: 0 – 8250
Graph:
A zero of a function
= solution of the equation
y(x) = 0
y is in this case 0, and the points that come out are all the points of the graph crossing the x-as.
In this case, only mention the outcomes of x! so x = …
Thus, y is always zero so no need to write it down.
Intersection point = the points where two graphs intersect
Elementary functions
1. Constant function: y(x) = c
- Always same answer
- Example: y(x) = 3 or y(x) = 5 1/3
- Graph: horizontal line crossing the y-as at one point
2. Linear function: y(x) = ax + b
- a≠0
- a = slope
- b crosses the y-as.
- Graph:
- Zero of a linear function:
solve y(x) = 0
-
- a = y 2 – y 1 / x 2 – x1
2
,- Example linear functions: y(x) = 5x + 3 and z(x) = -4x
a. Determine the zeros of y(x) and z(x).
So: y(x) = 0
5x + 3 = 0
5x = -3
x = -3/5
z(x) = 0
-4x = 0
x=0
b. Draw the graphs of y(x) and z(x).
Find two points and connect them.
c. Determine the intersection point of the graphs of y(x) and z(x).
y(x) = z(x)
5x + 3 = -4x
5x + 4x = -3
9x = -3
x = -3/9 = -1/3
y(-1/3) = 5 * -1/3 + 3 = 4/3
3. Quadratic functions: y(x) = ax2 + bx + c
- a≠0
- Graph:
a > 0, ∪
a < 0, ∩
- Example: y(x) = x2 – 5x + 2
a. Zeros of a quadratic function:
→ IMPORTANT!!!
x = 5 + √ or x = 5 - √
b. Graph:
a = 1, so ∪.
Already have the points on x-as, now find the point where x=0, so the y-as. This is c.
- Discriminant (D): b2 – 4ac
D > 0: 2 zeros → 2 points on x-as
D = 0: 1 zero → 1 point on the x-as
D < 0: no zero → no point on x-as
3
, - Example: y(x) = 2x2 + px + 4 ½. Determine the values of p such that:
a. y(x) has no zeros:
First: solve D = 0
D = b2 – 4ac
D = p2 – 4 * 2 * 4 ½
D = p2 – 36 = 0
p2 = 36
p = √36 = 6 or -6
So:
It must be: -6 < p < 6
b. y(x) has one zero:
D = p2 – 36
So p2 must be 36, so p = √36 = 6 or -6.
c. y(x) has two zeros:
D = p2 – 36
p = √36 = 6 or -6.
to get 2 zeros: p < -6 or p > 6
Solving inequalities
1. Set the inequality to zero. (Example: h(x) ≥ 0, h(x) > 0, h(x) ≤ 0, h(x) < 0).
2. Determine the zeros of h.
- Solve h(x) = 0
3. Determine the sign chart of h.
4. Find the solution set from the sign chart.
Example: Find all x such that 2x2 + 3x + 2 ≤ 4x + 3.
So:
1.
2x2 + 3x – 4x + 2 – 3 ≤ 0
h(x) = 2x2 – x – 1 ≤ 0
2.
h(x) = 0
4
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