How do you find the inverse function of f(x)=x^3+5f(x)=x3+5?

1 Answer
Sep 24, 2015

f^{-1}(x)=root(3)(x-5)f1(x)=3x5

Explanation:

Set y=f(x)=x^3+5y=f(x)=x3+5 and solve for xx: x^3=y-5\rightarrow x=root(3)(y-5)x3=y5x=3y5.

It's okay to write the answer as x=f^{-1}(y)=root(3)(y-5)x=f1(y)=3y5. It's also okay to swap variables and write the answer as y=f^{-1}(x)=root(3)(x-5)y=f1(x)=3x5.

Swapping the variables is a good idea when you want to graph both functions in the same picture while seeing the "reflection property" of a function and its inverse across the line y=xy=x. Not swapping the variables is a good idea if they have a real life meaning (which you would "lose" by swapping them).

Do note that the original function is indeed a one-to-one function since its graph passes the horizontal line test (see the graph below). There's no need to restrict its domain in any way.

graph{x^3+5 [-40, 40, -20, 20]}