In general the given function is in the form #y=f(x)# and its inverse is #f^(-1)(y)=x#, provided that #f(x)# is always crescent or decrescent for each value in the domain. This is just the case, being #f'(x)=5+11x^(10)# always positive for any #x in R#.

checked this, the problem of finding #f^(-1)(-13)=x# can be seen in an equivalent way as the one of finding the value of #x# whose image is #-13#.

To find it, it is enough to solve the equation #8x^(11)+5x-13=0#.

This has one root in #x=-1# so that at this point in the domain corresponds the value #y=-13# in the image. Being #f(-1)=13# and verified its monotonicity, we can concude that #f^(-1)(-13)=-1#