Math Is Fun Forum

Kiran

Member

Registered: 2006-11-15

Posts: 177

ok another important question

if they ask to find teh inverse function
and the eqaution looks like

y = number/numberX - number/number

now in my book..tehy are saying
a original function looks like this > y = 2x - 1
solved for x looks like this > x = 1/2y + 1/2
inverse function looks liek this > y = 1/2x + 1/2

now the problem that we have looks like an inverse function....
so now what?
so turn that problem around for it to look like :
x = number/numberY - number/number
right?

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mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: ok another important question

Saying 'number' lots of times like that gets really confusing. Is each 'number' a different constant?

I'm going to solve y = a/bx - c/d, where a, b, c and d are all constants. If that's not what you meant, then please explain in better detail.

The first thing we need to do is get all the terms involving x on their own. To do this, we move the c/d across to get a/bx = y + c/d.

Now we multiply by x to get it off the denominator: a/b = x(y + c/d).

Now we multiply by (y+c/d) to get x as the subject: x = [a/b]/[y + c/d].

While that fraction may look ugly, most of it is made up of constants, and we can define a/b and c/d as more constants, which makes it looks much nicer. x = e/(y+f), where e = a/b and f = c/d.

And once you've got it to this stage, you can switch x and y back to get the inverse function:
y = e/(x+f).

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It wanted to be normal.

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: ok another important question

x = 1/2y + 1/2
-in my book this is equally an inversefunction.
Swap x and y, however, doesn't change the nature of a function. In the function x = 1/2y + 1/2, the dependent variable is one half of the inverse of the independent variable plus one half.(dependent variable is the one on the left of the equater) In the function y = 1/2x + 1/2, the same thing stands. Since they could be called the same functions by nature, it is at least of some reason to say they are both the inverse functions.

Further, naming y = 1/2x + 1/2 the inverse function of y = 2x - 1 enjoys pratical benefits. Almost all the functions have the form y= some x, therefore we can merely figure out two inverse functions among the mass functions we encountered unless we define that way. Inverse functions defined in this way are symmantric to the line y=x. Hence in this way, we actually reduce the functions we need to learn by half, in the field of how to graph, for example.

In general, both ways are right, and the latter helps us a lot!

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Kiran

Member

Registered: 2006-11-15

Posts: 177

Re: ok another important question

ok now as i have it...

y = 1/3 - 5/6
i did what you said.
1/3x = y + 5/6
1/3= x(y + 5/6)
x = [1/3]/(y + 5/6)
now i think maybe you meant by this:
x = 0.3 / (y + 0.83)
correct?

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