Inverse Functions: nth Roots (Part 7)

In the previous posts of this series, I carefully considered the definition of f(x) = \sqrt{x} = x^{1/2}. Let’s now repeat this logic to consider the definition of f(x) = \sqrt[n]{x} = x^{1/n}, where n \ge 3 is an integer. We begin with n even.

A typical case is n =4; the graph of y = x^4 is shown below.


The full graph of y= x^n fails the horizontal line test if n is even. Therefore, we need to apply the same logic that we used earlier to define y = \sqrt[n]{x}. In particular, we essentially erase the left half of the graph. By restricting the domain to [0,\infty), we create a new function that does satisfy the horizontal line test, so that the graph of y = \sqrt[n]{x} is found by reflecting through the line y = x.

fourthroot2Written in sentence form,

If n is even, then y = \sqrt[n]{x} means that x = y^n and y \ge 0. In particular, this is impossible for real y if x < 0.

green line
We now turn to the case of n odd. Unlike before, the full graph of y= x^n (in thick blue) satisfies the horizontal line test. Therefore, there is no need to restrict the domain to define the inverse function. (shown in thin purple).



In other words,

If n is odd, then y = \sqrt[n]{x} means that x = y^n. There is no need to give a caveat on the possible values of y.

In particular, \sqrt{-8} and \sqrt[4]{-8} are both undefined since there is no (real) number x so that x^2 = -8 or x^4 = -8. However, \sqrt[3]{-8} is defined and is equal to -2 since (-2)^3 = -8.

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