Which Pair Of Functions Are Inverse Functions

How to Check Whether Two Functions Are Inverses of Each Other

Here is the method I use, which is basically what everyone uses once they get past the first week of algebra. You compose them. If f(g(x)) = x and g(f(x)) = x, they are inverses. That is the definition. The rest is bookkeeping. I run into this almost weekly when someone sends me a multiple-choice question with four pairs and asks which one works. The fastest way is to test each pair by composition rather than trying to invert one and match it. Let me walk through that. Take a pair like f(x) = 2x + 3 and g(x) = (x - 3)/2. I compute f(g(x)). That gives 2 * ((x - 3)/2) + 3. The 2s cancel, I get x - 3 + 3, which is x. Then I do g(f(x)): (2x + 3 - 3)/2 = 2x/2 = x. Both compositions return x, so these are inverses.

Now take a bad pair. f(x) = x^2 and g(x) = sqrt(x). People pick this one all the time because it looks right. It is not quite right without restrictions. f(g(x)) = (sqrt(x))^2 = x for x >= 0. That part checks out. But g(f(x)) = sqrt(x^2) = |x|, which is not x when x is negative. So they fail the second composition test unless the domain is explicitly restricted to non-negative numbers. This is the kind of trap that shows up on every standardized test and in my grading stack. Here is a realistic edge case I still see in student work: rational functions where the algebra looks clean but the domain gets weird. Consider f(x) = (x + 1)/(x - 2) and g(x) = (2x + 1)/(x - 1). At first glance, composing them is messy. I usually compute f(g(x)) by substituting and simplifying the compound fraction carefully. After clearing denominators, I check whether the result reduces to x. In this case it does, but I also have to note that x = 2 is excluded from f's domain and x = 1 is excluded from g's domain, and the composition f(g(x)) is undefined at x = 1. Beginners often skip that domain check and declare the pair correct anyway. When I catch that now, I just mark the missing domain restriction and move on. It costs me about two minutes per problem, but it prevents the wrong answer from looking right. The compositional test is reliable, but it has bottlenecks. If the functions involve absolute values, piecewise definitions, or trigonometric expressions, the algebra can take several minutes per pair. I sometimes switch to a graphical check when time is tight. I plot both functions on the same axes and look for symmetry across the line y = x. If one graph is the mirror image of the other across that diagonal, that is a strong hint. It is not a proof, though. I only use the graph as a quick filter before doing the composition math, especially when I am grading a long set of problems.

Another thing people get wrong is assuming that having an inverse means the original function is one-to-one everywhere. It does not. A function needs to be injective on its domain for an inverse to exist as a proper function. For example, f(x) = x^2 on all real numbers fails the horizontal line test, so it does not have an inverse function over its natural domain. If I restrict the domain to x >= 0, then the inverse is g(x) = sqrt(x), and the compositional test works cleanly. I remind students to check injectivity before they even start composing, because it saves them from chasing a dead end.

Which Pair Of Functions Are Inverse Functions

When you are given a list of pairs and asked to identify which pair of functions are inverse functions, the practical workflow is this. First, eliminate any pair where one function is clearly not one-to-one on the stated domain. Second, compute f(g(x)) and g(f(x)). If both equal x, the pair is correct. If only one does, the pair is incomplete or wrong. If the algebra is too long, use the y = x reflection check as a preliminary filter. I also recommend checking a single numeric point as a sanity test. Pick an x value inside the domain, evaluate f(x), then evaluate g at that output. You should get back to x. It catches sign errors and misplaced fractions quickly. For instance, with f(x) = 5x - 7 and g(x) = (x + 7)/5, I test x = 3. f(3) = 8, and g(8) = 3. That confirms the pair works numerically, and then I still do the formal composition to be thorough. There are cases where the compositional test succeeds but the domains are incompatible in a way that matters for the problem at hand. I ran into this recently with logarithmic and exponential pairs where the base was not e. f(x) = log_2(x) and g(x) = 2^x. The compositions give log_2(2^x) = x and 2^(log_2(x)) = x, so they pass. But if a question implicitly assumes a calculator-friendly base like 10 without stating it, and the answer choices mix bases, the pair still counts as correct mathematically even though the numeric evaluation path changes. I flag that mismatch when it appears in exam questions because it causes unnecessary confusion.

So in practice, to answer which pair of functions are inverse functions, you use composition, domain checks, and occasional numerical verification. The method is straightforward, the pitfalls are mostly about domain restrictions and one-to-one requirements, and you will catch most mistakes if you just verify both compositions and inspect the domain once you finish the algebra.