BAH! Domains & ranges

[Toothpaste]

Oh and here's something simple I found in my bound book that might be useful:

= Replace x in the equation with the number 2.

= Replace x in the equation with the expression for .

= Replace x in the equation with the value of

= Replace x in the equation with the number 2.

[/0]

Awesome stuff, thanks 

Did we indirectly answer your question too?

Yeah, I think so

[AppleXY]

I might also add to the existance of a composite function with high-res images illustrating it

[lanvins]

How do u find the domain of the following?: A rectangular piece of cardboard has dimensions 20cm by 36cm. Four squares each x cm by x cm are cut from the corners. An open box is formed by folding up the flaps

[phagist_]

well I'm not going to straight out tell you the answer =P

but... what are the rules (in terms of x) governing the length of each side (i.e from there look at the possible values x can take)

[Mao]

lol hehe i am

this is an "implied domain" question
as the equation is a model based on real life:
we cannot have 0 or negative width remaining, nor can we cut 0 or negative width

that constrains x to be (0,10)

[lanvins]

i don't get why it can't be (0,18)

[phagist_]

ok, how'd you arrive at those set of values?

I got 20-2x as the equation to one side and 36-2x as the equation to the other.

From 20-2x it has to be greater than zero.. as you cannot have a negative length...

[lanvins]

i got: 20-2x> 0,          36-2x>,                   x>0

which became:

-2x>-20,     -2x>-36,                   x>0

which became:
x<10,             x<18,                     x>0     

[Collin Li]

You have to take the intersection (overlap) of all your inequalities. The result is: (0,10)

(The reason why you take the intersection of them all, is because you need to satisfy all those conditions for the model to work)

[phagist_]

ahh I see... well what happens when you sub values from (0,18) into 20-2x>0 ?

It works fine from (0,10) but once you exceed that, you hit negative territory - hence it cannot work.

Try drawing a diagram and visualize what happens has x increases... once it approaches 10 one side becomes very very small, and should it exceed it, the side would have a negative length - which is impossible.

[lanvins]

This might be a dumb question but to phagist,um, how did you know to choose 20-2x>0 to sub into and not the others? thanks

[Collin Li]

If you're going to use maths, do what I said. Find all the inequalities, and find the intersection of them all.

You have to take the intersection (overlap) of all your inequalities. The result is: (0,10)

(The reason why you take the intersection of them all, is because you need to satisfy all those conditions for the model to work)

However, this is the "commonsense" way to think of it:

Code: [Select]

_________
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|_       _|
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If you have a rectangle: 20x36, then if the squares have a side length larger than 10, you'll run into troubles, because the squares will start to overlap each other. On the shorter side (the length 20 side), you'll have two squares of equal size that simply cannot be more than 10 cm each, or the squares will begin to overlap (10+10 = 20). What happens on the length 36 side does not matter, because that will allow a bigger square size, but we have already reasoned that any square size larger than 10 is incompatible with a rectangle that has a side length of 20.

[phagist_]

ok from your 2 equations 20-2x>0 and 36-2x>0 you solve both.

from 20-2x>0 you get x = (0,10) and from 36-2x>0 you get x = (0,18)
remember you have to satisfy both equations

so you know x = (0,10) satisfies both, what happens if we choose say, 13?

well it satisfies 36-2x>0, but hold on a sec it doesn't work with 20-2x>0 - it yields a negative answer, which we know it is impossible to have a negative length.

As I said draw it out and visualize what happens when x increases...

say x=9 (which satisfies both equations) you will now have a 2x18
you can see one side (the one governed by 20-2x>0) will reach zero before the other... anything after that (x=10,11,12,etc...) the side will have negative length (which not possible!) thus your rectangle is ruined.

as coblin said, you need to find the intersection of solutions so that there is a maximal domain which satisfies both equations.

[Mao]

ok from your 2 equations 20-2x>0 and 36-2x>0 you solve both.

from 20-2x>0 you get x = (0,10) and from 36-2x>0 you get x = (0,18)
remember you have to satisfy both equations

so you know x = (0,10) satisfies both, what happens if we choose say, 13?

well it satisfies 36-2x>0, but hold on a sec it doesn't work with 20-2x>0 - it yields a negative answer, which we know it is impossible to have a negative length.

As I said draw it out and visualize what happens when x increases...

say x=9 (which satisfies both equations) you will now have a 2x18
you can see one side (the one governed by 20-2x>0) will reach zero before the other... anything after that (x=10,11,12,etc...) the side will have negative length (which not possible!) thus your rectangle is ruined.

as coblin said, you need to find the intersection of solutions so that there is a maximal domain which satisfies both equations.

Lol
Moral of the day: when asked for implied domain, always use the shorter side