What set are you using S to represent?
In the attachment, regarding question 7, I had some confusion with whether or not to use the modulus function and why only the positive (1/x) is taken for 7b (or maybe it's only the negative, -1/x).
It would be fantastic if you could shed some light on this and clearly explain the solutions
Thanking you in advance
Sorry. When does the absolute value of 'x' replace just 'x' when squaring and square rooting?
Regarding question 7b, the f^-1(x) is 1/x^2
Then f(f^-1) = -1/sqrt(1/x^2) which equals 'x'. That means the sqrt taken was the negative 1/x not the positive. Why is this so?
Is this because of certain domain and range rules with inverse functions and functions of functions? What are they?
Thanks again
The OP is rightly alarmed that the usual swap-and-rearrange procedure results in
and hence
Which is a problem. So we are only allowed a negative in the expression of f inverse if it's inside the square, e.g.
whoops, maybe I'm confused, sorry if so... drawing a pic...
Ok, for my benefit anyway...
Thanks tom, I'm glad you understand; would you be able to explain your drawing please? I sort of understand but don't at the same time.
Does the absolute function come in to play at all in this question? (If so why is it? If not when does it?)
We have , but only for x ≥ 0, and for all real x.
Yes.
The left-hand side, f(f⁻¹), is not a complete expression: it lacks an x. Also, I don't see where -1 comes from in the right-hand side. We have . But if we consider the domain of f⁻¹ to be ℝ⁺ = {x ∈ ℝ | x > 0}, then f(f⁻¹(x)) = x.
This raises an important question: what is the domain of f⁻¹: ℝ - {0} or ℝ⁺? This depends on the definition of the inverse function, which I encourage you to double-check in your textbook. One option is to say that the domain of f⁻¹ is the range (i.e., image) of f. In this case, the range of f is ℝ⁺, which is a proper subset of the codomain ℝ. Then, as we've seen, f(f⁻¹(x)) = x for all x in the domain of f⁻¹.
In higher mathematics, though, the domain of the inverse function is usually defined as the codomain of the original function, i.e., ℝ in this case. Then f does not have an inverse because 1 / x² is not defined on 0. Even if we restrict the domain of f⁻¹ to ℝ - {0}, we would have f(f⁻¹(x)) = |x|, not x. According to this definition, in order to have an inverse, f must have its codomain declared as ℝ⁺, not ℝ.
Yes.
Incidentally, if we read as (which is unconventional but just as plausible in theory as a way of rendering as a function), then we don't run into the same problem with the natural algebraic formulation of the inverse.
Edit:
So no problem at all, of course.
Wow, I thought that this was an accidental pen stroke. This is bad typesetting. It should look like this: , or, better, like this: .
The inverse function is still , but its domain, as well as the range of f, is now ℝ⁻ = {x ∈ ℝ | x < 0}. This does not mean that we take the negative square root. We have because |x| = -x for x < 0.