In class, a student asked to work through a question:

Let f(x)=5(x1)(x+1)(x4)3x4.

(a) Show that f(x) can be written as 2x+1.

(b)Hence find f1(x), stating its domain.

The answer they gave was outrageous 1.

Part (a)

Part (a) was fine: combine it all into a single fraction as 5(x1)3(x+1)(x+1)(x4), then simplify the top to get 2(x4)(x+1)(x4)=2x+1. Not an enormous challenge for someone with algebraic fraction skills.

First part of (b)

… and the inverse function isn’t much more difficult: starting from x=2F(x)+1 (where F(x) is f1(x), but less hassle to write), we swiftly rearrange to get F(x)+1=2x and f1(x)=2x1, which can be written in several ways.

Second part of (b)

This is where the mark scheme and I disagree.

According to the mark scheme, the domain of the inverse function is xR, x0, and I have a problem with that.

And my problem goes back to part (a): the function is not properly defined: no domain is given. We can assume, reasonably, that it’s as big as the definition allows: all real numbers are allowed, except for -1 and 4.

Yes, and 4. Even though the ‘equivalent’ form, 2x+1 doesn’t have an x4 on the bottom, the function as it’s given to us does. Even though the limit of the function as x approaches 4 is well-defined (it gives 25), we still have to exclude it from the domain - we cannot evaluate 5(x1)(x+1)(x4)3x4 at x=4; neither term is defined.

And that has a knock-on effect for the inverse: because x=4 is excluded from the domain of f(x), f(x)=25 has to be excluded from its range. And if 25 is excluded from the range of f(x), it must also be excluded from the domain of f1(x).

So, according to me, the correct answer to the second part of (b) is xR, x0, x25.

Any objections? No. Good.

Footnotes:

1. Am I doing this clickbait thing right?