The question that's got me stuck is this: Find the value of (f

^{-1})'(x) for the given function (f) and the real number, a.

f(x)=(1/27)(x

^{5}+2x

^{3}), a=11

I know that (f

^{-1})'(x)= 1/(f'(f

^{-1})(x)) and all that, but it suddenly occurred to me when I went to solve this problem that I don't know how to find the inverse of the given function. It's one-to-one, so it seems like it must have one. But I only know how to solve for f

^{-1}(x) when the given function only has instances of x which are all of the same degree (or can be algebraically manipulated to be all the same degree) or when f(x) can be worked out to be the power of some binomial quantity (like I can solve for the inverse of x

^{2}+2x+1 because it can be compressed to (x+1)

^{2}. Every tutorial or help site I can find online about inverse functions offers only help on these kinds of functions.

I also found it odd that the derivative of the given function, f(x)=(1/27)(x

^{5}+2x

^{3}) (which is (1/27)(5x

^{4}+6x

^{2})), is not a 1-to-1 function, and should not, therefore, have an inverse. Is that part of the problem here? Thanks in advance for any help or advice.