# Inverse functions

1. Jun 6, 2005

### Struggling

find the inverse function of:
f: [-1/2,infin) ---> R, Where F(x) = 3sqaureroot 2x+1

i got f(x) = 3squareroot 2x+1
x = 3squareroot 2(f-1)+1
f-1 = (x^3-1)/2 where (infin,-1/2]

is this right?
because when i graph it, it doesnt really look like the inverse

also i just ran into another problem
find the inverse function of:
f : [-1,infin) ---R, where f(x) = x^2 + 2x - 8

i did:
f(x) = x^2 - 2x - 8
x = (f-1)^2 - 2(f-1) - 8
x+8 = (f-1)^2 - 2(f-1)
and then everything goes wrong here

edit: ARGGGGH just more and more problems, is there any forumla or way of working out the inverse of Log's and exponentials? i cant find any in my books

thanks

Last edited: Jun 6, 2005
2. Jun 6, 2005

### uart

No, nothing that you're doing there looks even close. You seem to really be at a loss to manipulate equations correctly. The basic skill you seem to be missing is that of transposing simple equations. Practice of this would be a good place for you to start.

For example, can you manipulate a simple equation like f = 9c/5 + 32 to express c in terms of f (this is called transposing btw). This is the thing you really need to lean before you can do the above problems (or just about anything else in maths actually).

3. Jun 6, 2005

### whozum

The inverse of the log function is the exponential.
the inverse of the exponential is the log function.

For example

$$log_3 y = x$$

Inverse function is $3^n$

$$3 ^ {log_3 y} = 3 ^ x [/itex] [tex] y = 3^x$$

4. Jun 6, 2005

### mathman

Rewrite x2-2x-8-f=0.

x=1+-sqrt(9+f)