SOLUTION: The function f defined by f(x) = x^5+x^3 is one-to-one (here the domain of f is the set of real numbers). Compute f^−1(y) for four different values of y of your choice. [

Algebra ->  Functions -> SOLUTION: The function f defined by f(x) = x^5+x^3 is one-to-one (here the domain of f is the set of real numbers). Compute f^−1(y) for four different values of y of your choice. [      Log On


   

Question 942371: The function f defined by
f(x) = x^5+x^3
is one-to-one (here the domain of f is the set of real
numbers). Compute f^−1(y) for four different values of y of your choice.
[For this particular function, it is not possible to find a
formula for f^−1(y).]
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
For example, if x = 2, then f(2) = 2^5 + 2^3 = 40. Then f^(-1) (40) = 2.

Do this for three other numbers.