London UK
Posts: 229 since Oct 2014

Any with an interest in science and technology interested in reigniting an interest in Mathematics here is a starter copy and paste from my post on another forum with further free starter informtion & material.
 Came across this actual problem in a book I chanced to glance through.
 I provide solution calculations (I have proof checked  however, accommodate any typos)  some basic calculus.
 Solutions verified against book answers  the book does not give calculations and the answeres can be derived by alternative methods.
The Question  5.2 Contextual Q:3:  On a particular day, the Financial Times 100 Share Index (FTSE 100) opens in London at 4,000.
 During the rest of the day, its value t hours at 9 a.m. is given as F = 4,000  16t^2 + 8t^3  3/4t^4.
 A broker is instructed to sell the client’s position (in preagreed increments and criteria – not relevant to Question and Solutions) only if the value of the FTSE is falling.
Comment: 16t^2 = 16 x t to the exponent 2; 8t^3 = 8 x t to the exponent 3; etc. (in first example  t is the base and 2 is the exponent and t^2 is the power)
Questions:
A: What is the value of the FTSE at noon.
B: Calculate the highest value of the index during the day and at what time to the nearest minute did this occur.
C: If trading finishes at 4.30 p.m., by how much has the index risen or fallen during the day.
D: During which times of the day could the broker have sold off the clients position ?
Answers:
A: 4011.3 points (note: indexes are designated in points (professionally) not pips)
B: 4183.9 points at 3.19 p.m.
C: Rise of 102.0 points at 4.30 p.m.
D: Between 9 a.m. and 10.41 a.m. and between 3.19 p.m. and 4.30 p.m.
SOLUTIONS:
Solution A:
12.00 (noon)  9.00 = 3. Therefore: t = 3. Note: 9.00 a.m. ~ t = 0; 10 a.m ~ t = 1; 11.00 a.m ~ t =2; 12.00 p.m. ~ t = 3.
At 9.00 a.m ~ t = 0
F = 4,000  16t^2 + 8t^3  3/4t^4
F= 4,000  16(0^2) + 8(0^3)  3/4(0^4)
F= 4,000  16(0) + 8(0)  3/4(0)
F= 4,000  0 + 0  0
The index = 4,000 points at 9.00 a.m. or when time t = 0.
At 12.00 ~ t = 3
F = 4,000  16t^2 + 8t^3  3/4t^4
F= 4,000  16(3^2) + 8(3^3)  0.75(3^4)
F = 4,000  16(9) + 8(27)  0.75(81)
F = 4,000  144 + 216  60.75
F= 4,011.25
F= 4,011.3
Answer A: The index vaue is 4,011.3 points at noon (12.00 p.m.)
Solution B:
See YouTube for basic differentiation
Differentiate F(t) to determine local and absolute maximum/minimum values; on a graph t(time) would be the horizontal axis and F (points) would be the vertical axis: normally the horizontal axis is designated the xaxis with xvalues and the vertical axis is designated yaxis with yvalues. In this instance t(time) is equivalent to the xaxis and index point values equivalent to the yaxis. > this may help:
Inserted Video
F = 4,000  16t^2 + 8t^3  3/4t^4
F'(t)  differentiated = 32t + 24t^2 3t^3 (note: F'(t) is differentiated to the first order; F"(t) is defferentiated to the second order, F"(t) = 32 + 48t  9t^2; third order, F'"(t) = 48 18t)
Note: F = 4,000  16t^2 + 8t^3  3/4t^4 can also be arranged as (pay attention to the  and + signs): F = 3/4t^4 + 8t^3  16t^2 + 4,000
and F'(t) = 32t + 24t^2 3t^3 can also be arrange: F'(t) = 3t^3 + 24t^2  32t The latter is the normal way  presenting expressions/equations in decending order of powers  but it is not necessary.
 It also become tedious to be pedantic about exact/strict/rigorous mathematical notation when knocking off a few quick solutions.
 However, whether decending or assending  incremental order of the powers should be maintained.
Factorise t: 32t + 24t^2 3t^3 = t(32 + 24t  3t^2)
a: t=0,
b: 32+24t3t^2 = 0
Apply quadratic formula to 32 + 24t  3t^2 or cubic formula to 32t + 24t^2 3t^3 (Sharp ELW505T, may be suitable, cheap and considered superior features to comparable Casio – and bonus one key press ease of access re Pi  (not referal links) ELW506T  Sharp calculatorsApplying quadratic formular to 32 + 24t  3t^2 (note: calculator inputs will be in order, 3,+24,32)
Then:
t2 = 1.690 (derived from factorisation > quadratic formular calculation) [calculator or pen/paper if hardcore]
t3 = 6.309 (derived from factorisation > quadratic formular calculation)
and t1 = 0 (derived from factorisation  refer a: )
Applying qubic formular to 32t + 24t^2  3t^3 (note: calculator inputs will be in order, 3,+24,32,0)
Then:
t1 = 0 (derived from cubic formula calculation) [calculator or pen/paper if hardcore]. Note: The 'solve any cubicsynthetic division method' as touted by clickbate YouTuber's is not univeral and only solve for the YouTuber's cherrypicked equations. For a fuller understanding of cubic formular* see  Further Information  link at end of post
t2 = 1.690 (derived from cubic formula calculation)
t3 = 6.309 (derived from cubic formula calculation)
Process t1
F1 = 4,000  16t1^2 + 8t1^3  3/4t1^4
F1 = 4,000  16(0^2) + 8(0^3)  4/4(0^4)
F1 = 4,000  0 + 0  0
F1 = 4,000 (point value at t = 0 or 9.00 a.m.)
Process t2
F2 = 4,000  16t2^2 + 8t2^3  3/4t2^4
F2 = 4,000  16(1.69^2) + 8(1.69^3) 3/4(1.69^4)
F2 = 4,000  16(2.8561) + 8(4.826809)  0.75 (8.15730721)
F2 = 4,000  45.6976 + 38.614472 6.117980408
F2 = 3,986.798892
F2 = 3,986.8 points
Calculating time: 1.69
Minutes: .69(60) = 41.4 = 41
Hours: 9.00 + 1 = 10
Time2 = 10.41
Process t3
F3 = 4,000  16t3^2 + 8t3^3  3/4t3^4
F3= 4,000  16(6.309^2) + 8(6.309^3) 3/4(6.309^4)
F3= 4,000 = 16(39.8034) + 8(251.120)  0.75(1,584.317)
F3 = 4,000  636.855 + 2,008.96  1,188.237
F3 = 4,183.868
F3 = 4,183.9
Calculation time: t = 6.309.
Hours: 9:00 + 6 = 15
Minutes: .309(60) = 18.54 = 19
Time3 = 15:19 = 3.19 p.m.
Answer B: The highest value of the index is 4,183.9 points at 3.19 p.m.
Solution C:
Time: t = 16.30  9.00 = 7.30 ; 30 minutes / 60 minutes = 0.5 ; t = 7.5
F = 4,000  16t^2 + 8t^3  3/4t^4
F = 4,000  16(7.5^2) + 8(7.5^3)  0.75(7.5^4)
F= 4,000  16(56.25) + 8(421.875)  0.75(3,164.0625)
F= 4,000  900 + 3375  2373.146875
F = 4,101.853125
F = 4,102
4,102  4,000 = 102
Answer C: The index has risen by 102 points at 4.30 p.m.
Solution D:
At 9.00 a.m. point value = 4,000.0
At 10:41 a.m. point value = 3,986.8 (this was the min value determined by 1st order calculus differentiation)
At 3.19 p.m. point value = 4,183.9 (this was the max valuedetermined by 1st order calculus defferentiation)
At 4.30 p.m point value = 4,102.0
The index was falling between 9.00 a.m and 10.41 a.m. and 3.19 p.m. and 4.30 p.m.
Answer D: The broker could have incrementally sold off the client's position between 9.00 a.m  10.41 a.m and 3.19 p.m. and 4.30 p.m.

Further Information  FREE Mathematical Refresher Material: 
I included this in the original post
Good joke – the nonsense should be fairly obvious. However, I wonder how obvious....
There is nothing special about today, this year, any other year or every 1,000 years.
If the age of anyone on the 31 December (note: the last day of the year) is added to the year of their birth the sum will >> always = the current year, exact.
If the age of anyone is calculated before their birthday of the current year then the sum of their age + year of birth will >> always = the current year  1 year (the current year less 1 year).
Example 1:
Anthony Hopkins (an actor, by way of example  see wikipedea, if interested) was born 31 Dec 1937.
This year, at current date  6 Dec 2022  Anthony is 84 years old.
84 + 1937 = 2021.
This year, on 31 Dec 2022  Anthony will be 85.
85 + 1937 = 2022. (the current year, exact)
Example 2:
Next year, on 1 Jan 2023 Anthony will  still  be 85.
85 + 1937 = 2022.
Next year, on the 31 Dec 2023 Anthony will be 86.
86 + 1937 = 2023. (the current year, exact)
