Inverse Functions: Square Root (Part 6)

In this post, I take a deeper look at the standard mistake of “simplifying” \sqrt{x^2} incorrectly as just x.

In the previous post, we noted that the function f: \mathbb{R} \to \mathbb{R} defined by f(x) = x^2 fails the horizontal line test and thus does not have an inverse function.



However, we can restrict the domain of f to make a new function that does satisfy the horizontal line test. This new function is identical to f where both f and g are defined. Following tradition, we restrict the domain to [0,\infty):

g: [0,\infty) \to [0,\infty) defined by g(x) = x^2.

So by essentially erasing the left half of the parabola, we form a function that passes the horizontal line test which therefore has an inverse. Naturally, this function is g^{-1}(x) = \sqrt{x}. When I teach Precalculus, I like to write this as a sentence:

y = \sqrt{x} means that x = y^2 and y \ge 0.



Since g and g^{-1} are inverse functions, it’s always true that g(g^{-1}(x)) = x and g^{-1}(g(x)) = x whenever these functions are defined. For example,

g(g^{-1}(16)) = g(\sqrt{16}) = g(4) = 4^2 = 16 and

g^{-1}(g(3)) = g^{-1}(3^2) = g^{-1}(9) = \sqrt{9} = 3

In other words, we are guaranteed that g^{-1}(g(x)) = g^{-1}(x^2) = \sqrt{x^2} is always equal to x — as long x lies in the domain of g… or, in other words, as long as x is nonnegative.

Because if x is negative, all bets are off.

Remember, the original function f does not have an inverse. In particular, g^{-1} and f are not inverse functions, and so it’s possible for g^{-1}(f(x)) to be something other than x. For example,

g^{-1}(f(3)) = \sqrt{3^2} = \sqrt{9} = 3, but

g^{-1}(f(-3)) = \sqrt{(-3)^2} = \sqrt{9} = 3 \ne -3

Of course, I don’t expect my Precalculus students to remember the subtle reason that this fails when they do their homework problems. But I do tell my Precalculus students that the nontrivial simplification of \sqrt{x^2} is a natural consequence of restricting the domain of a function that does not pass the horizontal line test to define an inverse function. In this example, \sqrt{x^2} = x as long as x \ge 0. However, if x < 0, then the result really could be just about anything else. For the current example, we have the pairwise simplification

\sqrt{x^2} = x if x \ge 0;

\sqrt{x^2} = -x if x < 0.

The last line is often uncomfortable for students, and so I remind them that x is assumed to be negative so that -x is positive. Of course, there’s a new notation that mathematicians have developed so that this two-line simplification of \sqrt{x^2} can be compressed to a single line:

\sqrt{x^2} = |x|

Unfortunately, in my opinion, this remains the problem that can be stated in two seconds or less (“Simplify the square root of x squared”) that, in my opinion, is missed most often by mathematics students.

One thought on “Inverse Functions: Square Root (Part 6)

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