Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**kynetikg01****Member**- Registered: 2006-01-24
- Posts: 3

hello everyone. i was doin some calculus and got stuck on this problem

f(x)= 10 - x

find the : f^(-1)(x)= ?

written as f'(x) just replace the ' w/ the -1

what does that mean? is that the inverse or somethIN?

i've been dealing w/ derivatives

-thanks

*Last edited by kynetikg01 (2006-02-28 17:12:13)*

Offline

**naturewild****Member**- Registered: 2005-12-04
- Posts: 30

I'm positive that it's asking for an inverse. But I dont know why you are dealing with inverse when you have learnt derivatives.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Ya it's inverse function.

Write it like this:

y=10-x

-x=y-10

x=10-y

This is the inverse function:

f^(-1)(y)=10-y

[corrected]

*Last edited by krassi_holmz (2006-03-01 04:30:18)*

IPBLE: Increasing Performance By Lowering Expectations.

Offline

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

I think there's a mistake in there.

y = 10-x

x+y = 10

x = 10-y

Therefore, f[sup]-1[/sup](x) = 10-x.

Why did the vector cross the road?

It wanted to be normal.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Mathsy's right.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

[corrected]

IPBLE: Increasing Performance By Lowering Expectations.

Offline

Pages: **1**