You have,Originally Posted by juangohan9
Switch the x and the y and solve,
Thus,
Thus,
Thus,
Thus,
Thus,
Hey, this might seem like baby work to some but I'm taking an online class hundreds of miles from campus, so I've got no one to turn to, but here.
So this is what I'm trying to find, the inverse of...
6x+8
f(x)= ----
28-x
In words, just in case the math gets screwed up,
F of x is equal to 6x+8 over/divided by 28-x... find the inverse...
I thought I had inverse of functions, but I cant get this one.
Thanks in advance for your help!
For ease of notation I'm going to call f(x) y. To get the inverse of y = f(x) we switch x and y to get x = f(y) and solve for y.Originally Posted by juangohan9
Thus:
Finally:
Thus your inverse function is .
You should, of course, check this by verifying and , as well as looking at domains and ranges.
-Dan
Actually, you need to show that an inverse function exists.Originally Posted by Quick
Then you need to show that the order is not important (mathematicians say "composition is commutative") meaning,
(1)
(2)
Since an inverse function exists I denote it by . Then I make condition (1) true.
Since,
I subsitute my inverse function (which is ) and have,
but I need it to give so I set it equal to .
---
But for high school problems composition is always commutative (you can ignore that) and the inversibility exists. So you can just procede with that step.
This will (formally) always work. Of course we might not be actually able to solve x = f(y) and we might find that the range of f(x) does not match the domain of the inverse function (and vice-versa). It may even be possible (I don't know how likely) that the domain of the inverse function winds up being empty due to a mismatch between the range of f(x) and the domain of the inverse.Originally Posted by Quick
-Dan
In general consider the composition of functions f(g(x)). In order for this to exist we need to have that the range of the function y = g(x) is the same as the domain of the function z = f(y), else the expression is meaningless. For example, take (range: [1, infinity) ) and (domain: (-infinity, 0] ). Then which has the empty set as a domain, and thus is a "function" that is defined nowhere.Originally Posted by Quick
-Dan
It depends on the notation. The "f" in f(x) represents an operator. As such it needs some sort of argument (ie. x) to make sense. However, we often abbreviate the notation to just "f" if the argument is not important. However, in an equation we need to know what the argument is so f=x+(1/x) is not a valid defining statement of a function. (For a variable f, yes, a function f(x), no.)Originally Posted by Quick
-Dan