one to one functions g and h are defined as follows. g={ (-9,7), (-1,-6), (1,4), (6,-1)} h (x)= 2x-9 and I have to find the following of: 1.) g^-1 (-1) = ? 2.) h^-1 (x) =? (h o h^-1) (1)= ? (I don't understand the steps to finding these answers..)
The first question is to find "g inverse of -1" which means you are looking for the number you would plug into g to get -1. Looking at how the function g works, it seems that the answer is 6, because of the point (6, -1). This can be interpreted as g(6) = -1, so g inverse of -1 must be 6.
so you are just switching the values since they are inverses?
so for the h^-1 (x) it would be 4?
sort of, basically, funtions and their inverses go "forwards" and "backwards" as an example, one of the points in g is (-9, 7), so g(-9) = 7, while g inverse of (7) = -9
ok i understand that but what am i supposed to do with the h (x)= 2x-9? I only use that when it asks a questions involving the h?
for that "h inverse of x" problem its a little bit trickier. you are looking for a rule for the inverse; another function. the easiest way i know how to do this is to solve for x. so:\[h(x) = 2x-9 \Longrightarrow h(x) +9 = 2x \Longrightarrow x = (h(x) +9)/2\]
Then for formalities, switch the x and h(x), and change the h(x) to h inverse of x
so i do the opposite of what it started out as?
yeah, your doing the opposite of the the function is telling you to do
btw the formal answer would be \[h^{-1}(x) = (x+9)/2\]
ok so what happens on the (h o h^-1) (1)= ? how do i figure out the h o h^-1 part... this part seems to be WAY more confusing
would it just be the answer without switching anything since it is reversing twice? or no?
Doing it by hand, that notation means you plug 1 into the inverse first and get an answer, then you plug that answer into the regular function, but since these functions are inverses of each other, it will spit out 1 as a final answer
notice, h inverse of 1 is (1 + 9)/2 = 5 then h(5) is 2(5) - 9 = 1
ok i think im understanding.. its just tricky. ill keep practicing with other problems now. THANK YOU
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