- 52

- 0

## Main Question or Discussion Point

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

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: