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The Funky function challange

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Stanley Boots

on 10 November 2016

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Transcript of The Funky function challange

The Five Functions:

A(x) = 4x - 3
B(x) = 3x²
C(x) = 7x + 1
D(x) = x²
E(x) = 2x + 7

How I will Organize the Functions
The first and most simple way to organize the functions, but you would have to guess and check many different arrangements of functions. It would not be the most logical way to solve the problem.
Solving A, E, and C
Which Arrangement of A, E, and C?
Now that we know all the possible outcomes of A, E, and C, which one is the best? The best one is the one with the biggest product.
A(C(E(2))) = 309
This is the arrangement of the A, E, and C with the biggest product, 309.

Question we are Answering:

What arrangement of functions will produce the largest output with two as as the input (each function used only once)? Two being the input number.

The Funky function challenge
By etoile Boots

More Logical Way to Solve the Problem:

So we know that guess and check is not a great to solve the problem, so how else can we solve it? Well, another way to solve it is to put functions that square a number at the end, in this case it is B and D. Why you may ask, this is because we all know that a number squared, will most likely be more than a number multiplied by another, especially since in this case, by the time that you apply B and D, the number will be very big, this will boost the number by 1,000s!
So what to do with A, C, and E?
Now that we know that we put A, E, and C first, and B and D last, how do we know which order to put A, E, and C? I will be using guess and check because there are only 6 possible ways that A, E and C can be organized, I will then choose which one has the largest product with 2 as the input number.
Solving A(C(E(2)))
1. E(2) = 2(2) + 7
E(2) = 4 + 7
E(2) = 11

2. C(11) = 7(11) + 1
C(11) = 77 + 1
C(11) = 78
3. A(78) = 4(78) - 3
A(78) = 312 - 3
A(78) 309
A(C(E(2))) =
309
ANSWER:
Solving A(E(C(2)))
1. C(2) = 7(2) + 1
C(2) = 14 + 1
C(2) = 15
2. E(15) = 2(15) + 7
E(15) = 30 + 7
E(15) = 37

3. A(37) = 4(37) - 3
A(37) = 148 - 3
A(37) = 145
ANSWER:
A(E(C(2))) = 145
Solving E(A(C(2)))
1. C(2) = 7(2) + 1
C(2) = 14 + 1
C(2) = 15
2. A(15) = 4(15) - 3
A(15) = 60 - 3
A(15) = 57
3. E(47) = 2(57) + 7
E(47) = 114 + 7
E(47) = 121
ANSWER:
E(A(C(2))) =
121
Solving E(C(A(2)))
1. A(2) = 4(2) - 3
A(2) = 8 - 3
A(2) = 5

2. C(5) = 7(5) + 1
C(5) = 35 + 1
C(5) = 36

E(36) = 2(36) + 7
E(36) = 72 + 7
E(36) = 79
ANSWER:
E(C(A(2))) = 79
Solving C(A(E(2)))
1. E(2) = 2(2) + 7
E(2) = 4 + 7
E(2) = 11
2. A(11) = 4(11) - 3
A(11) = 44 - 3
A(11) = 41

3. C(41) = 7(41) + 1
C(41) = 287 + 1
C(41) = 288
ANSWER:
C(A(E(2))) = 288
Solving C(E(A(2)))
1. A(2) = 4(2)- 3
A(2) = 8 - 3
A(2) = 5
2. E(5) = 2(5) + 7
E(5) = 10 + 7
E(5) = 17
3. C(17) = 7(17) + 1
C(17) = 119 + 1
C(17) = 120
ANSWER:
C(E(A(2))) = 120
What do we do with B & D?
Now we guess and check with B & D, there are only 2 possible outcomes so it will be very simple and easy.
Solving B and D
Solving B(D(309))
AKA B(D(A(C(E(2)))))
1. D(309) = 309²
D(309) = 95481

2. B(95481) = 3(95481²)
B(95481) = 3(9116621361
B(95481) = 27,349,864,083
ANSWER:
B(D(A(C(E(2))))) = 27,349,864,083
Solving D(B(309))
AKA D(B(A(C(E(2)))))
1. B(309) = 3(309²)
B(309) =3(95481)
B(309) = 286443
2. D(859329) = 286443²
D(859329) = 82,049,592,249
So which one?
We now know the solution for both B(D(A(C(E(2))))) and D(B(A(C(E(2))))). We found out that D(B(A(C(E(2))))) has a greater product, 82,049,592,249. So that is the order of functions that result the greatest solution.
D(B(A(C(E(2))))) =82,049,592,249
ANSWER:
Would your arrangement of functions necessarily be the best for all input (other than two) Explain why.
This arrangement of functions will be the best for all input number equal to or bigger than 0. I will explain it by showing some work of which arrangement of functions would be better for -5.
Solving A, E, and C for -5
Solving A(E(C(-5)))
1. C(-5) = 7(-5) + 1
-35 + 1
-34
2. E(-34) = 2(-34) + 7
-68 + 7
-61
3. A(-61) = 4(-61) - 3
-244 - 3
-247
Answer:
-247
Solving A(C(E(-5)))
1. E(-5) = 2(-5) + 7
-10 + 7
-3
2. C(-3) = 7(-3) + 1
-21 + 1
-20
3. A(-20) = 4(-20) - 3
-80 - 3
-83
Answer:
-83
Solving E(A(C(-5)))
1. C(-5) = 7(-5) + 1
-35 + 1
-34
2. A(-34) = 4(-34) - 3
-136 - 3
-139
3. E(-139) = 2(-139) + 7
-278 + 7
-271
Answer:
-271
Solving E(C(A(-5)))
1. A(-5) = 4(-5) - 3
-20 - 3
-23
2. C(-23) = 7(-23) + 1
-161 + 1
-160
3. E(-160) = 2(-160) + 7
-320 + 7
-313
Answer:
-313
Solving C(A(E(-5)))
1. E(-5) = 2(-5) + 7
-10 + 7
-3
2. A(-3) = 4(-3) - 3
-12 - 3
-15
3. C(-15) = 7(-15) + 1
-105 + 1
-104
Answer:
-104
Solving C(E(A(-5)))
1. A(-5) = 4(-5) - 3
-20 - 3
-23
2. E(-23) = 2(-23) + 7
-46 + 7
-39
3. C(-39) = 7(-39) + 1
-273 + 1
-272
Answer:
-272
So Which Arrangement of A, C, and E for -5?
So you can see from my calculations that A(C(E(-5))) has the greatest value, but since in the end, the negative number will be squared, it will become a positive number, so in this case, we want use the arrangement of functions that has the smallest value, because it is the greater negative number, if you understand what I mean. The function combination is E(C(A(-5))). This proves that D(B(A(C(E())))) does not work for all numbers, only for 0 and above.
Find another input number for which a different order of the functions would produce the largest output? What would be the order of functions to produce the largest output for this input number?
Challenge Question:
Challange Question:
We have already found a different input number that needs a different arragement of functions, -5. We also know the arrangemnt of E, A, and C already so we just need to figure out B and D.
Solving B and D
Solving B(D(-313))
aka. B(D(E(C(A(-5)))))
1. D(-313) = -313²
97969
2. B(97969) = 3(97969²)
3(9597924961)
28,793,774,883
Answer:
28,793,774,883
Solving D(B(-313))
aka. D(B(E(C(A(-5)))))
1. B(-313) = 3(-313²)
3(97969)
293907
2. D(293907) = 293907²
86,381,324,649
Answer:
86,381,324,649
Solution
The solution for what arrangement of functions that will have the greatest value with the input number of -5 is D(B(E(C(A(-5))))) This arrangement of functions would work for all negative numbers.
Conclusion:
My conclusion is that you need different arrangements for negative and positive numbers. For postive numbers, or numbers over 0, the arrangement is D(B(A(C(E(+#))))). For negative numbers though, the arrangement of functions goes like this, D(B(E(C(A(-#))))).
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