It looks to me like the student was following the argument about how to find the inverse of a function alright, but requires remedial training in how to solve an equation for a variable. The student does not appear to understand how the order of operations of squaring a variable, then adding three to it, must be undone in reverse order.

Michael Paul Goldenberg

I’m having trouble in the first example distinguishing between what the student wrote and what (perhaps) someone else did commenting. That said, I like to use arrow diagrams to teach inverse functions. So for f(x) = 2x – 3, given input x, multiply by 2, then subtract 3 from the result to get the output. The inverse function, if it exists, could then be found by reversing the direction (order) of the arrows, and their operators. So given that output (2x – 3), add 3 (yielding 2x) and then divide by 2, yielding x, the original input. This shows that the inverse function of 2x – 3 = (x +3)/2, which is not what is given and not what the student did. Neither did s/he try an example: putting a value for x into f(x), then putting the output number back into the proposed (and incorrect) inverse to see what popped out.

My sense is that the student figures that order of operations doesn’t matter if ignoring them “helps your cause.” That is, assume that anything goes if it will lead you to “the inverse.” ;^)

In the second example, domain and range seem not to matter to the student. The various tests for the existence of an inverse function (horizontal line test, odd v. even function, etc.) don’t seem to matter. So it doesn’t bother the student that taking sqr(x^2) must give TWO answers for positive inputs (and none for negative inputs in the real numbers). My guess is that at this point, the notion of “inverse function” is very poorly understood by this student. A lot of grasping at straws: if something looks plausibly close to what an inverse function might “smell like” for a given function, “go for it.”

I’m also noticing that the student seems to want to apply operators to the variable but not so much tot he constant. In the first example 2x is divided by 2 but the -3 is not. In the second example the square root operation is applied only to the x^2 term, not the – 3 term.

I’ve always wondered why we call the inverse of f(x) to be g(x) when the inputs have to be different kinds of things in general. I bet we’d see a lot fewer student errors if we called y = f(x) and then asked for the inverse function g(y).

Having them keep an ordered pair (x, f(x)) and then testing that the inverse function does the right thing with it might also be a good thing to do to remind them of what the inverse really is about and check the answer.

I agree that the first student solution looks like they understand the general idea of what’s going on and the second looks more like a problem in equation-solving than in the concept of inverses.