Given #f(x)= x^5+x^3+x#, how do you find the inverse f(3) and f(f inverse (2))?

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Answer 1 · 2015-10-18T07:59:54

Examine the behaviour of #f(x)# and of #f^(-1)# to find:

#f^(-1)(3) = 1# and #f(f^(-1)(2)) = 2#

#f(x) = x^5+x^3+x# is a strictly monotonically increasing function with domain and range the whole of #RR#.

One way of seeing that is to look at its derivative:

#f'(x) = 5x^4 + 3x^2 + 1 > 0# for all #x in RR#

So its inverse #f^(-1)# is also strictly monotonic increasing with domain and range the whole of #RR#.

#f^(-1)(f(x)) = x# for all #x in RR#

Applying #f# to both sides we get:

#f(f^(-1)(f(x))) = f(x)# for all #x in RR#

Substitute #y = f(x)# to find:

#f(f^(-1)(y)) = y# for all #y in RR#

This last "for all #y in RR#" is due to the range of #f(x)# being the whole of #RR#.

So #f# is the inverse of #f^-1# too.

For example, #f(f^(-1)(2)) = 2#

Also, if #f(x) = 3#, then #f^(-1)(3) = x#

#f^(-1)(3)# is the root of #f(x) = 3#

That is:

#x^5+x^3+x = 3#

Notice that #x = 1# works, so this will be the only Real root and

#f^(-1)(3) = 1#