Inverse of a function

**juangohan9** · August 3rd 2006, 10:27 AM

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!

**ThePerfectHacker** · August 3rd 2006, 10:34 AM

Originally Posted by juangohan9

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!

You have,
$\frac{6x+8}{28-x}$
Switch the x and the y and solve,
$\frac{6y+8}{28-y}=x$
Thus,
$6y+8=(28-y)x$
Thus,
$6y+8=28x-xy$
Thus,
$6y+xy=28x-8$
Thus,
$y(6+x)=28x-8$
Thus,
$y=\frac{28x-8}{x+6}$

**Quick** · August 3rd 2006, 10:36 AM

Originally Posted by ThePerfectHacker

Switch the x and the y and solve,

Is that how to solve inverse functions for all problems or is this a special case...

**topsquark** · August 3rd 2006, 10:38 AM

Originally Posted by juangohan9

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.

Thus: $x = \frac{6y+8}{28-y}$

$x(28-y) = 6y+8$
$28x - xy = 6y +8$
$28x -8 = 6y + xy$
$28x -8 = y(6+x)$
Finally:
$y = \frac{28x - 8}{x+6} = 4 \frac{7x - 2}{x+6}$

Thus your inverse function is $f^{-1}(x) = 4 \frac{7x - 2}{x+6}$ .

You should, of course, check this by verifying $f \left ( f^{-1}(x) \right ) = 1$ and $f^{-1}(f(x)) = 1$ , as well as looking at domains and ranges.

-Dan

**ThePerfectHacker** · August 3rd 2006, 10:40 AM

Originally Posted by Quick

Is that how to solve inverse functions for all problems or is this a special case...

Actually, you need to show that an inverse function exists.
Then you need to show that the order is not important (mathematicians say "composition is commutative") meaning,
$f(f^{-1}(x))=x$ (1)
$f^{-1}(f(x))=x$ (2)
Since an inverse function exists I denote it by $y$ . Then I make condition (1) true.
Since,
$f(x)=<br /> \frac{6x+8}{28-x}$
I subsitute my inverse function (which is $y$ ) and have,
$\frac{6y+8}{28-y}$ but I need it to give $x$ so I set it equal to $x$ .
---
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.

**ThePerfectHacker** · August 3rd 2006, 10:43 AM

Originally Posted by topsquark

You should, of course, check this by verifying $f \left ( f^{-1}(x) \right ) = 1$ and $f^{-1}(f(x)) = 1$ , as well as looking at domains and ranges.

No, you need to show,
$f \left ( f^{-1}(x) \right ) = x$ and $f^{-1}(f(x)) = x$

Because the function $f(x)=x$ is the identity element for composition.

**topsquark** · August 3rd 2006, 10:44 AM

Originally Posted by Quick

Is that how to solve inverse functions for all problems or is this a special case...

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.

-Dan

**topsquark** · August 3rd 2006, 10:45 AM

Originally Posted by ThePerfectHacker

No, you need to show,
$f \left ( f^{-1}(x) \right ) = x$ and $f^{-1}(f(x)) = x$

Because the function $f(x)=x$ is the identity element for composition.

Oops!

I was thinking of my Algebra book: $f \left ( f^{-1}(x) \right ) = I$ where "I" is the identity function. My apologies!

-Dan

**topsquark** · August 3rd 2006, 10:46 AM

Wow! Lots of overlapping posts on this one.

-Dan

**Quick** · August 3rd 2006, 10:49 AM

Originally Posted by topsquark

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?.

-Dan

Another question: if we have equation $f(x)=x^2+1$ could we say that f=x+(1/x) or do we alwayse have to keep f and x together?

**ThePerfectHacker** · August 3rd 2006, 10:53 AM

Originally Posted by Quick

Another question: if we have equation $f(x)=x^2+1$ could we say that f=x+(1/x) or do we alwayse have to keep f and x together?

Not all functions has inverses

This one does not.

An informal test, is to pass the horizontal line and see if it intersects in two places.

**topsquark** · August 3rd 2006, 10:54 AM

Originally Posted by Quick

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 $g(x) = x^2 + 1$ (range: [1, infinity) ) and $f(y) = ln(-y)$ (domain: (-infinity, 0] ). Then $f(g(x)) = ln(-x^2 - 1)$ which has the empty set as a domain, and thus is a "function" that is defined nowhere.

-Dan

**topsquark** · August 3rd 2006, 10:57 AM

Originally Posted by Quick

Another question: if we have equation $f(x)=x^2+1$ could we say that f=x+(1/x) or do we alwayse have to keep f and x together?

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.)

-Dan

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