SOLUTION: In reference to Inverse Functions: How do you determine when to use the "short cut" method when trying to find the inverse of a function? For example, f(x) y = (x3+8)/3 so, we cu

Algebra ->  Functions -> SOLUTION: In reference to Inverse Functions: How do you determine when to use the "short cut" method when trying to find the inverse of a function? For example, f(x) y = (x3+8)/3 so, we cu      Log On


   

Question 15349: In reference to Inverse Functions: How do you determine when to use the "short cut" method when trying to find the inverse of a function? For example, f(x) y = (x3+8)/3 so, we cubed it, added 8, and then divided by 3. If you undo it in reverse order (multiply 3, minus 8, cube root) it becomes the inverse. f%5E%28-1%29%28x%29+=+root%283%2C3x-8%29 Why won't this work on all the problems? Thank you in advance!
Found 2 solutions by khwang, rapaljer:
Answer by khwang(438) About Me  (Show Source):
You can put this solution on YOUR website!
You are right for the inverse of
f(x)= y = %28x%5E3%2B8%29%2F3
to get
y = +f%5E%28-1%29%28x%29+=+%283x-8%29%5E%281%2F3%29+

But use ^ for exponent and parentehsis for grouping terms next time.
If the original function is constructed by elementary invertible
function (such as addition, multiplication,...) , we only need to
apply reverse process step by step to get the inverse function.

Since y=f(x)= T%5Bk%5DT%5Bk-1%5D..T%5B1%5D%28x%29 then
f%5E%28-1%29%28x%29 = T%5B1%5D%5E%28-1%29 T%5B2%5D%5E%28-1%29..T%5Bk%5D%5E%28-1%29%28x%29
(if each T%5Bi%5D+ is invertible.]
By for some functions such as y = x/(x^2+1) (or the functions containing
some transcendental functions as trig or log )
is not so easy to get 1st inverse. I suggest that you try.

Anyway, you have got good idea and good luck!!!
Kenny

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
I'll just BET that this question was raised by one of MY own students, with reference to #17 on page 329 of my College Algebra: One Step at a Time!!! Am I right??? Congratulations on a GREAT question, and thanks to Kenny for a great answer. "Invertible" steps! That's what you call it.

R^2 at Seminole Community College
Sanford, Florida