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

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.

]]>How can you do that?]]>

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.

]]>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.]]>

y = S(x+L)/x²

yx² = Sx + SL

yx² - Sx - SL = 0

Using quadratic equation:

x = (S ± √(S² - 4ySL) ) / 2y

Which is the inverse.

]]>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.

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