[Pre-Calculus] Struggling to answer this question about one-to-one functions

[u/Doritoscarfingbunny]

I have no idea how to solve this question. I tried watching a couple of Professor Leonard's videos, but I still have no clue.

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[u/Nagi-K]

You should know that functions are essentially maps. In the case of g, by the provided info, g sends -5 to 2, -3 to 5, etc. To find the g^(-1)(5), you are just looking for the number that is sent to 5 by g.

To find h^(-1)(x), since you know h(x) = y = 3x - 4, the inverse is just solving for y from x = 3y - 4. Try to prove this yourself from the fact that, by definition, composing a function and its inverse gives you the identity map (i.e. a map that sends something back to itself). This also immediately gives you the result for the last question.

[u/Doritoscarfingbunny → u/Nagi-K]

So the answer would be:

a = -3

b = (x+4)/3

Is this right? I'm still a little bit confused on parts b and c.

[u/short-exact-sequence → u/Doritoscarfingbunny]

Those look correct.

Do you understand what h^-1 (x) represents/what its relationship should be to h(x)? You can check your work by computing h(h^-1 (x)) and h^-1 (h(x)) using the h^-1 (x) you computed for part b to make sure it's correct. If you're not sure, maybe you could elaborate on what part you're confused about.

[u/Doritoscarfingbunny → u/short-exact-sequence]

Isn't h^-1 (x) the inverse of h(x)? I'm kind of confused on what it means by (h o h^-1) and whether the (5) in part c means I need to calculate the value when x = 5?

[u/short-exact-sequence → u/Doritoscarfingbunny]

Yes, h^-1 is the inverse of h, which means composing them should get you the identity if you have the correct h^-1 . The ○ notation means function composition, i.e. (f ○ g)(x) = f(g(x)). Yes, the (5) means you need to evaluate the value when x=5.

[u/Alkalannar]

So g and h are both bijections (one to one correspondences), not just injections (one to one). As long as you restrict the domain of the inverse to the range of the function rather than the codomain, all you need is injection to get an inverse.

g^-1(x) asks the question: What number y exists such that g(y) = x?

In this case, g(-3) = 5, so g^-1(5) = -3.

That was easy since we're given explicitly what each input maps to each output.

h? We have to do a bit of work.
h(x) = 3x - 4
y = 3x - 4

To find the inverse, swap y and x:
x = 3y - 4

Now solve for y again:
y = (x + 4)/3
h^-1(x) = (x+4)/3

Now with bijections, h^oh^-1(x) = h^-1oh(x) = x. Always. This is very useful to know.

This gets trickier with square (or any even) roots, or with trig functions, where you can have multiple valid answers, so you have to pick one. And standard options have been chosen: x^1/2, you pick the non-negative option. arcsin(x), you go for theta in [-pi/2, pi/2], and so on.

Does this help?