# Finding the Inverse of a Funtion

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• November 5th 2008, 02:34 PM
Finding the Inverse of a Funtion
http://i255.photobucket.com/albums/h...hdude/math.jpg

How would I go about solving this?
• November 5th 2008, 03:14 PM
Luke774
First make it a "Y=" equation.

Y = x2-5x+6

Switch the Y with the x's

X = Y2-5Y+6

Solve for X

X-6 = Y2-5Y

X-6/5 = Y2

Square root of ((x-6)/5)) = Y

Kind of confusing, but I hope this helps!
• November 5th 2008, 06:01 PM
euclid2
Quote:

http://i255.photobucket.com/albums/h...hdude/math.jpg

How would I go about solving this?

The easiest way to go abouts doing this is making a table of values for the function and switching the values of the y and x. what i mean is make the x values the y values and the y values the x values. This will give you the inverse, f-1(x)
• November 6th 2008, 03:29 AM
HallsofIvy
Quote:

You don't. That is not a "one to one" function and so does not have an inverse. If $f^{-1}(x)$ is the inverse of f(x), then we must have $f^{-1}(f(x))= x$, that is, $f^{-1}$ must 'undo' whatever f does. But f(6)= 0 and f(-1)= 0. $f^{-1}(0)$ can't be both 6 and -1!
Luke774, for some reason, chose to take only the "+" sign on the square root. That's valid but gives the inverse of a slightly different function: $f(x)= x^2- 5x- 6$ with $x\ge 5/2$ and undefined for x< 5/2. Same formula but different domain so different function.
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