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**yonski****Member**- Registered: 2005-12-14
- Posts: 67

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.

Student: "What's a corollary?"

Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Why did the vector cross the road?

It wanted to be normal.

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**yonski****Member**- Registered: 2005-12-14
- Posts: 67

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

Student: "What's a corollary?"

Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**yonski****Member**- Registered: 2005-12-14
- Posts: 67

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

Student: "What's a corollary?"

Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,579

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)*

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,579

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

How can you do that?

**igloo** **myrtilles** **fourmis**

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**austin81****Member**- Registered: 2005-03-21
- Posts: 39

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

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