function machines

• February 14th 2009, 09:23 AM
andyboy179
function machines
this is my question Attachment 10116

i need to know what will be in the first box so all the answers = 3,6,10 if anyone can help i will be very greatfull thanks
• February 14th 2009, 10:55 AM
HallsofIvy
Quote:

Originally Posted by andyboy179
this is my question Attachment 10116

i need to know what will be in the first box so all the answers = 3,6,10 if anyone can help i will be very greatfull thanks

Let "f" represent the function in that first box. Since you then add 2 you are looking for a function so that
f(4)+ 2= 3, f(12)+ 2= 6, and f(28)+ 2= 10.

That is the same as saying that f(4)= 1, f(12)= 4, and f(28)= 8. You can, at worst, fit a parabola through three points so try f(x)= $ax^2+ bx+ c$. Putting in the given values, $f(4)= 16a+ 4b+ c= 1$, $f(12)= 144a+ 12b+ c= 10$, and $f(28)= 784a+ 28b+ 1= 10$. That gives you three equations to solve for a, b, c.

There are, of course, an infinite number of correct solutions to this problem.
• February 14th 2009, 10:59 AM
andyboy179
Quote:

Originally Posted by HallsofIvy
Let "f" represent the function in that first box. Since you then add 2 you are looking for a function so that
f(4)+ 2= 3, f(12)+ 2= 6, and f(28)+ 2= 10.

That is the same as saying that f(4)= 1, f(12)= 4, and f(28)= 8. You can, at worst, fit a parabola through three points so try f(x)= $ax^2+ bx+ c$. Putting in the given values, $f(4)= 16a+ 4b+ c= 1$, $f(12)= 144a+ 12b+ c= 10$, and $f(28)= 784a+ 28b+ 1= 10$. That gives you three equations to solve for a, b, c.

There are, of course, an infinite number of correct solutions to this problem.

thanks for what you just posted but i forgot to say the number in the first box can only be 1 number i can't have a load of different numbers
• February 14th 2009, 01:19 PM
stapel
Quote:

Originally Posted by andyboy179
i forgot to say the number in the first box can only be 1 number i can't have a load of different numbers

So the box can contain only one operation...? Then:

You have:

. . . . .x = 4: [something done with 4] + 2 = 3
. . . . .x = 12: [something done with 12] + 2 = 6
. . . . .x = 28: [something done with 28] + 2 = 10

Getting rid of the "plus two" after the "something done", and labelling the black-box output as "y", you have:

. . . . .x = 4: y = 1
. . . . .x = 12: y = 4
. . . . .x = 28: y = 8

For the first one, the "something done" might be "subtract three", but this won't work with the other two. And no subtraction that works for the either of the other two will work for the first. Clearly, addition won't work for any of them.

You would multiply in the first one by 1/4 (or, which is the same thing, divide by 4), but this won't work for the others, and nothing that works for either of the others will work for the first.

This shows that the only way to find a rule that works is to use the more complicated stuff that the previous replies explained. So either the instructions (to do one simple step using only one other number) were wrong, or else there is something wrong with the inputs or outputs.

Please consult with your instructor regarding corrections and clarification. Thank you! :D