- #1
Struggling
- 52
- 0
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 doesn't 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 can't 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 doesn't 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 can't find any in my books
thanks
Last edited: