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Topic review (newest first)
- austin81
- 2005-12-24 01:14:03
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
- Ricky
- 2005-12-22 13:03:40
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.
- John E. Franklin
- 2005-12-22 10:13:40
x and y are variables, so I don't think you can set y equal to a.
How can you do that?
- Ricky
- 2005-12-22 02:32:50
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 15:37:33
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.
- yonski
- 2005-12-15 07:35:32
It was staring at me all along! Thank you very much 
- Ricky
- 2005-12-15 07:30:48
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.
- yonski
- 2005-12-15 06:45:32
Okay cool thanks. It's really annoying me too now lol!
- mathsyperson
- 2005-12-15 06:14:38
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
- 2005-12-15 05:51:22
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.