Finding an inverse function

yonski · 2005-12-14 06:51:22

Hey,
i'm struggling with finding inverse functions so I was wondering whether anyone would be able to run me through it? For simple ones I can manage it, but this one is really bugging me:

f(x) = S(x+L)/x^2

I've plotted it on my calculator, so I know that the domain would have to be restricted to make the inverse function valid. But no matter how I rearrange it I just can't figure it out Do I have to do some differentiation or something?

Any help would be greatly appreciated!

Jon.

mathsyperson · 2005-12-14 07:14:38

It's got me stumped too. I've never seen a problem involving finding the inverse that needed differentiation before, and I don't think this one's any different. It's just quite hard. I'll try it again later, because it's annoying me with its unsolved-ness.

yonski · 2005-12-14 07:45:32

Okay cool thanks. It's really annoying me too now lol!

Ricky · 2005-12-14 08:30:48

S(x+L) is S*(x+L) and not a function, right? Using y as f(x) (as it's easier):

y = S(x+L)/x²
yx² = Sx + SL
yx² - Sx - SL = 0

Using quadratic equation:

x = (S ± √(S² - 4ySL) ) / 2y

Which is the inverse.

Last edited by Ricky (2005-12-14 08:31:16)

yonski · 2005-12-14 08:35:32

It was staring at me all along! Thank you very much

John E. Franklin · 2005-12-20 16:37:33

Ricky, I thought you have to set y=0 for quadratic equation. Please explain.
Plus, what is the inverse of a function, is it when you swap x and y axis and look
at the graph through the back of the paper, or just look through the back of the
paper and turn 90 degrees?? I can't remember.
...
Okay, I looked up inverses, (x,y) goes to (y,x), so look through back of
paper after flip paper through the y=x slope of 1 axis.
But I still don't understand the use of the quadratic equation with y as a variable and
not y set to zero.

Last edited by John E. Franklin (2005-12-20 16:44:21)

Ricky · 2005-12-21 03:32:50

If you draw the line y = x, the inverse of a function should be mirrored across that line. That is, f(x,y) = f-¹(y,x)

But I still don't understand the use of the quadratic equation with y as a variable and
not y set to zero.

y isn't the variable. x is:

yx² - Sx - SL = 0

a = y, b = -S, c = -SL

ax² + bx + c = 0

There are two ways to find the inverse. Take y = f(x), and swap y with x, then solve for y. This is more natural to most students because they are used to having y as the dependant variable and solving for it.

I prefer to take the other route. That is, keep x and y in the same place and just solve for x. You will find the same exact inverse, only x will be the dependant variable instead of y.

John E. Franklin · 2005-12-21 11:13:40

x and y are variables, so I don't think you can set y equal to a.
How can you do that?

Ricky · 2005-12-21 14:03:40

Variable or constant, it doesn't matter, a is just a "label" for y.

For example:

y = x^2 + 2

z = y

z = x^2 + 2

You really aren't saying anything new, just calling y a different name.

austin81 · 2005-12-23 02:14:03

S(x+L) is S*(x+L) and not a function, right? Using y as f(x) (as it's easier):

y = S(x+L)/x²
yx² = Sx + SL
yx² - Sx - SL = 0

Using quadratic equation:

x = (S ± √(S² - 4ySL) ) / 2y, y sholud not be 0.
That is ok now, i think

Math Is Fun Forum

#1 2005-12-14 06:51:22

Finding an inverse function

#2 2005-12-14 07:14:38

Re: Finding an inverse function

#3 2005-12-14 07:45:32

Re: Finding an inverse function

#4 2005-12-14 08:30:48

Re: Finding an inverse function

#5 2005-12-14 08:35:32

Re: Finding an inverse function

#6 2005-12-20 16:37:33

Re: Finding an inverse function

#7 2005-12-21 03:32:50

Re: Finding an inverse function

#8 2005-12-21 11:13:40

Re: Finding an inverse function

#9 2005-12-21 14:03:40

Re: Finding an inverse function

#10 2005-12-23 02:14:03

Re: Finding an inverse function

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