Think of a function as a math machine with an input and an output. Suppose the function is A(x) = 3x + 1. That means if you put any number (x) into this function machine, the machine will multiply the number by three and add one. For example, what is the output if five is the input? Five times three plus one equals 16 for the output. That's written A(5) = 16. Suppose you join two function machines so that the output of the first one is connected to the input of the second one. Let's make the second function B(x) = x^2, (remember that x^2 means x squared). If you put two into the A machine, out comes seven. Then seven is the input to the B machine which squares it, and out comes 49. Okay? That's written B(A(2)) = 49.
Here are five functions:
A(x) = 4x - 3
B(x) = 3x^2
C(x) = 7x + 1
D(x) = x^2
E(x) = 2x + 7
I am going to connect them in alphabetical order and use two as the input number. That would be written as E(D(C(B(A(2))))). The resulting output = 553,359.
What arrangement of functions will produce the largest output with two as the input (each function used only once)? I am looking for logical reasoning and algebra rather than just guess and check. Explain why you chose your particular arrangement. (Just write the order of your functions for the short answer.)
When it came to solving this POW you should have noticed there were two types of equations present linear and exponential. Knowing that eponentials grow so rapidly you would want to use their power as the end of the string of equations and the linears up front. You would test the class of equations to see which would produce the highest yeild in its group and order accordingly.
How did you do?
These are the five functions:
ReplyDeleteA(x) = 4x - 3
B(x) = 3x^2
C(x) = 7x + 1
D(x) = x^2
E(x) = 2x + 7
C.A.
This is my order:
C(x) = 7x + 1
E(x) = 2x + 7
A(x) = 4x – 3
B(x) = 3x^2
D(x) = x^2
Since you have to make the biggest outcome, I would start with the function with the biggest numbers, then go down. So I would do: C, E, A, B, D. I chose C first because C has the number 7, and then it adds one. Then I chose E because it multiplies the number by 2 and then it adds 7. Then for A, it is multiplying by 4 then subtracting 3. B is multiplying by 3 and then taking it to the power of 2. Then D is taking it to the power of 2. I think this is the most logical because you are starting out big, then getting smaller. My answer is: C, E, A, B, D.
JS
ReplyDelete2(2)+7=11
4(11)-3=41
7(41)+1=288
288^2=82,944
3(82,944)^2=20,639,121,408
I put the functions that have exponents in them last because exponents normally get a higher product than just normal multiplication. I put 3x^2 last because it will get a higher answer than x^2. Then I put the other function in that order greatest answer to least answer with the number 2. I put them in greatest to least because it would get the highest answer in the end.
MNH
ReplyDelete2(2)+7=11
4(11)-3=41
7(41)+1=288
288^2=82,944
3(82,944)^2=20,639,121,408
I put the functions that have exponents in them last because exponents normally get a higher product than just normal multiplication. I put 3x^2 last because it will get a higher answer than x^2. Then I put the other function in that order greatest answer to least answer with the number 2. I put them in greatest to least because it would get the highest answer in the end.
Five Functions:
ReplyDeleteA(x) = 4x - 3
B(x) = 3x^2
C(x) = 7x + 1
D(x) = x^2
E(x) = 2x + 7
My order:
E(x) = 2x + 7
A(x) = 4x – 3
C(x) = 7x + 1
D(x) = x^2
B(x) = 3x^2
M.D.
Since it is all about order and the greatest output, I would say that the functions for B and D would be the last two in the order, because they both involve exponents, which would give you a higher output than just multiplication. Since B is Bx and then squared, where D is just x squared, I would say that B would be the very last function in the order, because it would give a higher output. That would leave D as the second to last function, which means the order so far is D), B). Next, I would say that the third to last function in the order would be C, because in C you multiply x by 7 and then add 1. I decided this because after I used B and D, I knew I had A, C, and E left. A is 4x – 3, C is 7x + 1, and E is 2x + 7. Even though in E you add 7, in C you multiply it by 7, and then add 1, which is more than just 2x + 7. So, that would leave my order so far as C), D), B). After that, I knew that I had A and E left. I knew that in A you multiplied by 4 and subtracted 3. Even though you subtract 3, you still multiply it by 4, where as in E you multiply it by 2 and add 7. Based on that, I decided E would go first, and A would be second. That would leave my final order as E), A), C), D), B).
2x + 7 = 11 x 4 – 3 = 41 x 7 + 1 = 288 x 288 = 82944 x 82944 x3 = 20639121408
ReplyDeleteReasons for putting the exponents last because exponents usually get a lot higher outcome then multiplication. I wanted the machines that gave bigger outcomes last because when there 1st they don’t give a higher number then with a bigger number for example 8 squared is going to be a lot less then 16 squared. So I use all the machines that got lower outcomes in the beginning so they could raise the number then that number could get squared
E(A(C(D(B(2))))) the resulting output equals 20639121408.
ReplyDeleteReason for putting the exponents last is that if you multiply a high number by itself it would be bigger than multiplying the number by 2. So I had the machines that had a small outcome in the beginning so the ending numbers from the small numbers could be multiplied by itself and would make it a very large number. H.H
2x + 7 = 11 x 4 – 3 = 41 x 7 + 1 = 288 x 288 = 82944 x 82944 x3 = 20639121408
ReplyDeleteReasons for putting the exponents last because exponents usually get a lot higher outcome then multiplication. I wanted the machines that gave bigger putcomes last because when there 1st they don’t give a higher number then with a bigger number for example 8 squared is going to be a lot less then 16 squared. So I use all the machines that got lower outcomes in the beginning so they could raise the number then that number could get squared
S.A
HB
ReplyDeleteORDER: E(A(C(D(B(2)))))
2x2+7=11
4x11-3=41
7x41+1=288
288^2=82,994
3x82,994^2=20,639,121,408
The reason I chose this order is because I looked at all the equations and found the equation that would turn out the highest number. I also looked that exponents would turn out a higher number then any of the regular multiplication in the other problems. So for my last one it was the highest turn out then the other exponent. After that I looked at the multiplication problems and found the highest to lowest outcome for them. Then I ordered it in my answer from the highest outcome for the closest to last number then the rest.
JD
ReplyDeleteE(A(C(D(B(2)))))
2x2+7=11
4x11-3=41
7x41+1=288
288^2=82,994
3x82,944^2=20,639,121,408
I put the functions with the exponents first because an equation with exponents usual get a higher answer than an equation without an exponent. And the higher the number the higher the answer. So I put the two with exponents last. Then I ordered the functions from least to greatest for the answers with just the number 2. That got me the largest number.
My Order
ReplyDeleteA(x) = 4x – 3
C(x) = 7x + 1
E(x) = 2x + 7
D(x) = x^2
B(x) = 3x^2
In this problem, the last equations should be able to give you the largest amount, so that means the first equation should give you the least amount. Since the A equation subtracts an amount it should be the first equation in the order. It was a hard decision whether to put equation C or E as the second equation. But, equation C would give the higher amount because the X is multiplied by 7, rather than in equation E where it’s only multiplied by 2. So far, the value would be 79. The next equation would be equation D, because it is one of the equations with an exponent, but with the least computation. So that means that the last equation would be equation B, 3x^2. It multiplies the previous value by 3 first, and then squares that amount, which would give you the greatest amount. The final amount would be 350,550,729.
----A.M.
The first step I made was plugging in the number two for each equation and finding the outcome.
ReplyDeleteIf the input is two ..
A(x) = 4x – 3 Equals 5
B(x) = 3x^2 Equals 12
C(x) = 7x + 1 Equals 15
D(x) = x^2 Equals 4
E(x) = 2x + 7 Equals 11
The order I would put it in is C, B, E, A, and D.
The reason that would be my order is because the outcomes are largest to smallest. If we kept on going, which means that any other larger number (in each one of them) will come out with a larger outcome.
Also another reasoning would be is that I also put equations C & D at the beginning and end is not just because of their outcomes but because they are both linear equations and linear equation can increase very quickly which is why on graphs it makes a slope.
First, I plugged two into each equation. Then I took the functions
ReplyDeletethat created larger numbers, and I knew that I needed to put the
larger numbers in those functions. Knowing this, I put those equations last. They went last because by the time the numbers got to those equations they would be large numbers, hint, hint, wink, wink.
If the input is two…
A(x) = 4x – 3 equals_5
B(x) = 3x^2 equals_12
C(x) = 7x + 1 equals_15
D(x) = x^2 equals_4
E(x) = 2x + 7 equals_11
Therefore, the order would be C, B, E, A, AND D.
2
^
C(x) = 7x + 1 = 11
11
^
B(x) = 3x^2 = 363
363
^
E(x) = 2x + 7 = 726
726
^
A(x) = 4x – 3 = 2904
2904
^
D(x) = x^2 = 8433216
P.O.W
ReplyDeleteA(x) = 4x – 3
B(x) = 3x^2
C(x) = 7x + 1
D(x) = x ^ 2
E(x) = 2x + 7
My order:
A(x) = 4x – 3
C(x) = 7x + 1
E(x) = 2x + 7
D(x) = x ^ 2
B(x) = 3x^ 2
The reason I chose this order is because I started with the equation that would give me the least amount, and made it so the last equation would give me the most amount.
If we use two as x, A(x) = 4x – 3 would equal 5. In C(x) = 7x + 1 (x would equal 5) so it would equal 36. In E(x) = 2x + 7 (x would equal 36) so it would equal 79. In D(x) = x ^ 2
(x would equal 79) so 6,241, and in B(x) = 3x^2 (x would be 6,241) so the final answer would be 350,550,729.
In my order, I wanted to put the equation that would give me the most result at the end, and the least at the top. The one that gave me the most was equation B, the one that gave me the least was A. In the middle I had trouble deciding if I should put equation C or E first, C would gave me the second least amount, so I put equation C first, and E second. I put D after C because it gave me the second most amounts. So my order of equations is A, C, E, D, B.
-J.V
N.H
ReplyDeleteEACDBwould probably make the biggest out of the number two. The number that it produces is 74863626544. It has the largest number of digits.
E(x) = 2x + 7
A(x) = 4x – 3
C(x) = 7x + 1
D(x) = x^2
B(x) = 3x^2
We looked at the equations and found what equations, with the input of 2, would make the highest output to the lowest output. Our arrangement from highest to lowest was ADECB. We saw that ADECB didn’t not make the highest output compared to the arrangements we did previously. Then we decided to do highest to lowest and see if that would make a difference. It made a difference but it still wasn’t the highest. So far, the highest is E,D,C,B,A. We then did all functions using the input 2. Then we put the functions in order from lowest ton highest. That came out to be 484,416. It still wasn’t the highest.
ReplyDeleteSo our answer is EDCBA which is 25, 887,741
1. A,B,C,D,E
2
5
225
1576
2,483,776
2. E,D,C,B,A
11
121
848
6471936
25,887,741
3. ADECB
5
25
57
400
1,440,000
4.BCEDA
36
253
513
263169
1,052,673
5. D,A,E,C,B
4
13
33
232
484416
MO and MW
ReplyDeleteWe looked at the equations and found what equations, with the input of 2, would make the highest output to the lowest output. Our arrangement from highest to lowest was ADECB. We saw that ADECB didn’t not make the highest output compared to the arrangements we did previously. Then we decided to do highest to lowest and see if that would make a difference. It made a difference but it still wasn’t the highest. So far, the highest is E,D,C,B,A. We then did all functions using the input 2. Then we put the functions in order from lowest ton highest. That came out to be 484,416. It still wasn’t the highest.
So our answer is EDCBA which is 25, 887,741
1. A,B,C,D,E
2
5
225
1576
2,483,776
2. E,D,C,B,A
11
121
848
6471936
25,887,741
3. ADECB
5
25
57
400
1,440,000
4.BCEDA
36
253
513
263169
1,052,673
5. D,A,E,C,B
4
13
33
232
484416
P.M.
ReplyDeleteCADBE. The order is the smallest to biggest because 11 x 5 is bigger than 4 x 5
JHS
ReplyDeleteA(x) = 4x – 3
B(x) = 3x^2
C(x) = 7x + 1
D(x) = x^2
E(x) = 2x + 7
B E C A D
INPUT NUMBER: 2
B(2)3^2=36
E(36)2+7=79
C(79)7+1=554
A(554)4-3=2213
D(2213)^2=4,897,369
Our first idea was to put one of the exponential equations in the beginning to get the highest number starting out, and one in the end to finish off with a the biggest possible number. For the in-between linear equations we used all possibilities and used all the in-between equations and found the one with the most which was BECAD.
17363069361 caebd
ReplyDelete7*2-3=15
4*2-3=57
57*2+7=121
(3*121)^2=13179
13179^2=17363069361
I found this answer by multiplying from largest to smallest answer and then I squared the rest of the numbers.
S.D.
E.D.
ReplyDeleteA(2)= 4X-3
C(2.5)= 7x+1
B(7.4)= 3X^2
D(22.2)= X^2
E(51.4)= 2X+7=109.8
I chose this arrangement because you get the largest output this way.
A (3) = 4x-3 9
ReplyDeleteB (9) = 3x^2 729
C (729) = 7x + 1 5104
D (5104) = x^2 26,050,816
E (26050816) = 2x + 7 52,101,639
I chose this combination because it goes into the millions.
B E C A D
ReplyDeleteB- (2)3^2=36
E- (36)2+7=79
C- (79)7+1=554
A- (554)4-3=2213
D- (2213)^2=4,897,369
B E C A D
ReplyDelete(2)3^2 = 36
(36)2+7 = 79
(79)7+1= 554
(554)4-3= 2213
(2213) ^ 2 = 4 897 396
E.I.D.
ReplyDeleteA(2)= 4x-3
C(2.5)= 7x+1
B(7.4)= 3x^2
D(22.2)= x^2
E(51.4)= 2x+7=109.8
I chose this arrangement because it gives you the largest output. I chose to use 2 as my input.
BS
ReplyDeleteAt first I put the exponent ones last because I wanted to get the biggest number I could before I did the big final stretch with exponents.
C=15
A=57
E=121
B=43923
D=1929229929
C=15
A=57
E=121
D=14641
B=643076643
Then I started putting an exponent in the middle and at the end.
C=15
A=57
B=9747
E=19501
D=380289001
C=15
E=37
D=1369
A=5473
B=59907458
But I found that after trying lots of combos, I had the right idea first. My biggest number was 1929229929. The combo was CAEBD.
J.H
ReplyDeleteBDCAE
1. 36
2. 1296
3. 9073
4. 36289
5. 72585
We chose this because we wanted to do the exponential equations first and then we went to the other largest ones.
S.G.
ReplyDeleteTry 3: C (5) = 7x = 35 + 1 = 36 / C = 7.2
D (7.2) = x^2 = 51.84 / D = 7.2
E (7.2) = 2x = 14.4 + 7 = 21.4 / E = 2.9
A (2.9) = 4x = 11.6 – 3 = 8.6 / A = 2.9
B (2.9) = 3x = 8.7^2 = 75.69 / B = 26.1
I think that this was the best thing to go with in order to get the highest number possible, because I did three trials, and this one came out with the highest number.
Think of a function as a math machine with an input and an output. Suppose the function is A(x) = 3x + 1. That means if you put any number (x) into this function machine, the machine will multiply the number by three and add one. For example, what is the output if five is the input? Five times three plus one equals 16 for the output. That's written A(5) = 16. Suppose you join two function machines so that the output of the first one is connected to the input of the second one. Let's make the second function B(x) = x^2, (remember that x^2 means x squared). If you put two into the A machine, out comes seven. Then seven is the input to the B machine which squares it, and out comes 49. Okay? That's written B(A(2)) = 49.
ReplyDeleteHere are five functions:
A(x) = 4x – 3 = 5
B(x) = 3x^2 = 12
C(x) = 7x + 1 = 15
D(x) = x^2 = 4
E(x) = 2x + 7 = 11
I am going to connect them in alphabetical order and use two as the input number. That would be written as E(D(C(B(A(2))))). The resulting output = 553,359.
What arrangement of functions will produce the largest output with two as the input (each function used only once)? I am looking for logical reasoning and algebra rather than just guess and check. Explain why you chose your particular arrangement. (Just write the order of your functions for the short answer.)
1 b = 12 1c = 15 1 c 7x + 1=15
2 d = 144 2a = 57 2 a 4(15)-3 = 57
3 a = 573 3e = 121 3 b 3x^2 = 9747
4 e = 1153 4b = 43923 4 e 2x + 7 = 95004016
5 c = 8072 5d = 1929229929 5 d x^2 = 9025763056128
AB
ReplyDeleteC, B, E, A, D….The greatest to the lowest original outcome
• If I used 2 as the fist input.
• Sample equations are the a, b, c, and d equations using 2 as the permanent input.
1. 15 7*2+1
2. 675 3*15^2
3. 1357 2*675+7
4. 5425 4*1357-3
5. 29,430,625 5425^2
The combination that I think gives you the highest outcome is CBEAD. This is because on the “sample equations,” I ordered them from the greatest to the least original outcomes.
Equation: x^2(4x-3(2x+7(3x^2(7x+1(x)))))